tt     . 


NEW 


ELEMENTARY  ALGEBRA: 


EMBRACING 


THE  FIRST  PRINCIPLES  OF  THE  SCIENCE. 


PKOFE880K    OF    HIGHER    MATH  KM  ATIO8,    COLUMBIA    COLLEOB. 


NEW  YORK  : 
BARNES  &  BURR,  PUBLISHERS,  51  &  53  JOHN  ST. 

CHICAGO:    GEORGE   SHERWOOD.   118  LAKE   ST. 

CINCINNATI:  RICKEY  AND  CARROLL. 

ST.  LOUIS :  KEITH  AND  WOODS. 

1804. 


A.    S.    BARNES     AND     BUHIi'S     PUBLICATIONS. 

Da  vies1    Course    of   Mathematics. 

MATHEMATICAL    WORKS, 

IS  A  SERIES  OF  THREE   PAKT8  : 

ARITHMETICAL,     ACADEMICAL,     AND     COLLEGIATE 


DAVIES'  LOGIC  AND  UTILITY  OF  MATHEMATICS. 
Tins  series,  combining  all  that  is  most  valuable  in  the  various  methods  of  Enropeas 
Instruction,  improved  and  matured  by  the  suggestions  of  more  than  thirty  years' 
experience,  now  forms  the  only  complete  consecutive  course  of  Mathematics.  Ita 
methods,  harmonizing  as  the  works  of  one  mind,  carry  the  student  onward  by  the 
same  analogies,  and  the  same  laws  of  association,  and  are  calculated  to  impart  a  com- 
prehensive knowledge  of  the  science,  combining  clearness  in  the  several  branches,  and 
unity  and  proportion  in  the  whole  ;  being  the  system  so  long  in  use  at  West  Point, 
through  which  so  many  men,  eminent  for  their  scientific  attainments,  have  passed, 
and  having  been  adopted,  as  Text  Books,  by  most  of  the  Colleges  in  the  United  States. 

I.— THE     ARITHMETICAL     COURSE     FOR     SCHOOLS. 

1.  PRIMARY   ARITHMETIC   AND   TABLE-BOOK. 

2.  INTELLECTUAL   ARITHMETIC. 

3.  SCHOOL  ARITHMETIC.     (Key  separate.) 

4.  GRAMMAR    OF    ARITHMETIC. 

II.— THE     ACADEMIC     COURSE. 

1.  THE  UNIVERSITY  ARITHMETIC.     (Key  separate.) 

2.  PRACTICAL   MATHEMATICS   FOR   PRACTICAL   MEN. 

3.  ELEMENTARY  ALGEBRA.     (Key  separate.) 

4.  ELEMENTARY  GEOMETRY  AND  TRIGONOMETRY. 

5.  ELEMENTS  OF  SURVEYING. 

III.— THE   COLLEGIATE   COURSE. 

1.  DAVIES'  BOURDON'S  ALGEBRA. 

2.  DAVIES'  UNIVERSITY  ALGEBRA. 

3.  DAVIES'  LEGENDRE'S  GEOMETRY  AND  TRIGONOMETRY. 

4.  DAVLES'  ANALYTICAL  GEOMETRY. 

5.  DAVIES'  DESCRIPTIVE  GEOMETRY. 

6.  DAVIES'  SHADES,  SHADOWS,  AND  PERSPECTIVE. 

7.  DAVIES'  DIFFERENTIAL  AND  INTEGRAL  CALCULUS. 

8.  MATHEMATICAL  DICTIONARY,  BY  DAVIES  &  PECK. 

ENTEBED,  according  to  the  Act  of  Congress,  in  the  year  one  thousand  eight  hundred 
and  fifty-nine,  by  CHAELES  DAVIES,  in  the  .Clerk's  Oflice  of  the  District  Court  of  the 
United  States,  for  the  Southern  District  of  New  York. 

1TILI.IAM   DEXYSE,   STEREOTYPES. 

a 


P  K  E  F  A  C  E. 


ALGEBRA  naturally  follows  Arithmetic  in  a  course  of  scien- 
tilic  studies.  The  language  of  figures,  and  the  elementary 
combinations  of  numbers,  are  acquired  at  an  early  age. 
When  the  pupil  passes  to  a  new  system,  conducted  by 
letters  and  signs,  the  change  seems  abrupt;  and  he  often 
experiences  much  difficulty  before  perceiving  that  Algebra 
is  but  Arithmetic  written  in  a  different  language. 

It  is  the  design  of  this  work  to  supply  a  connecting  link 
between  Arithmetic  and  Algebra ;  to  indicate  the  unity  of 
the  methods,  and  to  conduct  the  pupil  from  the  arithmetical 
processes  to  the  more  abstract  methods  of  analysis,  by  easy 
and  simple  gradations.  The  work  is  also  introductory  to 
the  University  Algebra,  and  to  the  Algebra  of  M.  Bourdon, 
which  is  justly  considered,  both  in  this  country  and  in 
Europe,  as  the  best  text-book  on  the  subject,  which  has  yet 
appeared. 

In  the  Introduction,  or  Mental  Exercises,  the  language 
of  figures  and  letters  are  both  employed.  Each  Lesson  is 
so  arranged  as  to  introduce  a  single  principle,  not  known 

iii 


IV  PREFACE. 

before,  and  the  whole  is  so  combined  as  to  prepare  the 
pupil,  by  a  thorough  system  of  mental  training,  for  those 
processes  of  reasoning  which  are  peculiar  to  the  algebraic 
analysis. 

It  is  about  twenty  years  since  the  first  publication  of  the 
ELEMKNTAEY  ALGEBRA.  Within  that  time,  great  changes 
have  taken  place  in  the  schools  of  the  country.  The  sys- 
tems of  mathematical  instruction  have  been  improved,  new 
methods  have  been  developed,  and  these  require  correspond- 
ing modifications  in  the  text-books.  Those  modifications 
have  now  been  made,  and  this  work  will  be  permanent  in 
its  present  form. 

Many  changes  have  been  made  in  the  present  edition,  at 
the  suggestion  of  teachers  who  have  used  the  work,  and 
favored  me  with  their  opinions,  both  of  its  defects  and 
merits.  I  take  this  opportunity  of  thanking  them  for  the 
valuable  aid  they  have  rendered  me.  The  criticisms  of 
those  engaged  in  the  daily  business  of  teaching  are  invalu- 
able to  an  author ;  and  I  shall  feel  myself  under  special 
obligation  to  all  who  will  be  at  the  trouble  to  communicate, 
to  me,  at  any  time,  such  changes,  either  in  methods  or  lan- 
guage, as  their  experience  may  point  out.  It  is  only  through 
the  cordial  co-operation  of  teachers  and  authors — by  joint 
labors  and  mutual  efforts — that  the  text-books  of  the  country 
can  be  brought  to  any  reasonable  degree  of  perfection. 

COLUMBIA  COLLEGE,  NEW  YORK,  March,  1859. 


CONTENTS. 


CHAPTER      I. 

DEFINITIONS  AND  EXPLANATORY  SIGHS. 

PAQM. 

Algebra — Definitions — Explanation  of  the  Signs 83-41 

Examples  in  writing  Algebraic  expressions 41 

Interpretation  of  Algebraic  language 42 

CHAPTER    II. 

FUNDAMENTAL    OPERATIONS. 

Addition — Rule — Examples 48-50 

Subtraction — Rule — Examples — Remarks 60-56 

Multiplication — Monomials — Polynomials 56-63 

Division — Monomials 63-68 

Signification  of  the  symbol  «° 68-70 

Division  of  Polynomials — Examples 71-76 

CHAPTER    III. 

USEFUL   FORMULAS.      FACTORING,    ETC. 

Formulaa  (1),  (2),  (3),  (4),  (5),  and  (6) 76-79 

Factoring • 79-81 

Greatest  Common  Divisor 81-84 

Least  Common  Multiple 84-87 

CHAPTER    IV 

FRACTIONS. 

Transformation  of  Fractions  .' 89 

fo  Reduce  an  Entire  Quantity  to  a  Fractional  Form 90 

V  ^ 


VI  CONTENTS. 


To  Reduce  a  Fraction  to  its  Lowest  Terms 

To  Reduce  a  Fraction  to  a  Mixed  Quantity 

To  Reduce  a  Mixed  Quantity  to  a  Fraction 93 

To  Reduce  Fractions  to  a  Common  Denominator 94 

Addition  of  Fractions 96-98 

Subtraction  of  Fractions 98-99 

Multiplication  of  Fractions 99-102 

Division  of  Fractions 102-105 

CHAPTER    V. 

EQUATIONS  OF  THE  FIRST  DEGRIB. 

Definition  of  an  Equation — Different  Kinds 105-106 

Transformation  ot  Equations — First  and  Second 106-110 

Solution  of  Equations — Rule 110-1 14 

Problems  involving  Equations  of  the  First  Degree 115-130 

Equations  involving  Two  Unknown  Quantities 130-131 

Elimination — By  Addition — By  Subtraction — By  Comparison. .  131-143 

Problems  involving  Two  Unknown  Quantities 143-148 

Equations  involving  Three  or  more  Unknown  Quantities 148-159 

CHAPTER    VI. 

FORMATION   OF  POWERS. 

Definition  of  Powers 160-161 

Powers  of  Monomials 161-163 

Powers  of  Fractions 163-165 

Powers  of  Binomials 1 65-1 67 

Of  the  Terms— Exponents — Coefficients 167-170 

Binomial  Formula — Examples 170-172 

CHAPTER    VII. 

SQUARE  ROOT.      RADICALS  OF  THE  SECOND  DEGREE. 

Definition — Perfect  Squares — Ride — Examples 173-179 

Square  Root  of  Fractions 179-181 

Square  Root  of  Monomials 181-183 

Imperfect  Squares,  or  Radicals 1 83-1 87 

Addition  of  Radicals '. 187-189 

Subtraction  of  Radicals  ..  189-190 


CONTENTS.  Vll 


Multiplication  of  Radicals  ...............................  190-191 

Division  of  Radicals  ....................................  191-192 

Square  Root  of  Polynomials  .............................  193-197 

CHAPTER    VIII. 

EQUATIONS    OF  THE  SECOND  DEGREE. 

Equations  of  the  Second  Degree  —  Definition  —  Form  .........  198-200 

Incomplete  Equations  ____  .  ..............................  200-209 

Complete  Equations—  Rule  ...............................  209-211 

Four  Forms  ...........................................  21  1-227 

Four  Properties  ........................................  227-229 

formation  of  Equations  of  the  Second  Degree  ..............  229-231 

Numerical  Values  of  the  Roots  ...........................  231-236 

Problems  ..................  ............................  236-240 

Equations  involving  more  than  One  Unknown  Quantity  .  .....  241—250 

Problems   .............................................  250-254 

CHAPTER    IX. 

ARITHMETICAL  AND  GEOMETRICAL  PROPORTION. 

Ways  in  which  Two  Quantities  may  be  Compared  ...........  255 

Arithmetical  Proportion  and  Progression  ..................  256-257 

Last  Term  .........................  .  ...................  257-260 

Sum  of  the  Extremes—  Sum  of  Series  .....................  260-262 

The  Five  Numbers—  To  find  any  number  of  Means  ..........  262-265 

Geometrical  Proportion  .................................  267 

Various  Kinds  of  Proportion  ...................  .  .........  268-278 

Geometrical  Progression  .................................  278-280 

Last  Term—  Sum  of  Series  ...............................  280-285 

Progression  having  an  Infinite  Number  of  Terms  ............  285-288 

The  Five  Numbers—  To  find  One  Mean  ....................  288-289 

CHAPTER    X. 

LOGARITHMS. 

Theory  of  Logarithms  ...................................  290-291 


SUGGESTIONS    TO    TEACHERS. 


1.  THE  Introduction  is  designed  as  a  mental  exercise.     If 
thoroughly  taught,  it  "will  train  and  prepare  the  mind  of 
the  pupil  for  those  higher  processes  of  reasoning,  which  it 
is  the  peculiar  province  of  the  algebraic  analysis  to  develop. 

2.  The  statement  of  each  question  should  be  made,  and 
every  step  in  the  solution  gone  through  with,  without  the 
aid  of  a  slate  or  black-board  ;  though  perhaps,  in  the  begin- 
ning, some  aid  may  be  necessary  to  those  unaccustomed  to 
such  exercises. 

3.  Great  care  must  be  taken  to  have  every  principle  on 
which  the  statement  depends,  carefully  analyzed ;  and  equal 
care  is  necessary  to  have  every  step  in  the  solution  distinctly 
explained. 

4.  The  reasoning  process  is  the  logical  connection  of  dis- 
tinct apprehensions,  and  the  deduction  of  the  consequences 
which  follow  from  such  a  connection.     Hence,  the  basis  of 
all  reasoning  must  lie  in  distinct  elementary  ideas. 

5.  Therefore,  to  teach  one  thing  at  a  time — to  teach  that 
thing  well — to  explain  its  connections  with  other  things, 
and  the  consequences  which  follow  from  such  connexions, 
would  seem  to  embrace  the  whole  art  of  instruction. 

via 


ELEMENTARY   ALGEBRA. 


INTRODUCTION. 

MENTAL      EXERCISES. 

LESSON    I. 

1.  JOHN  and  Charles  have  the  same  number  of  apples ; 
both  together  have  twelve :  how  many  has  each  ? 

ANALYSIS.— Let  x  denote  the  number  which  John  has ; 
then,  since  they  have  an  equal  number,  x  will  also  denote 
the  number  which  Charles  has,  and  twice  x,  or  2aj,  will 
denote  the  number  which  both  have,  which  is  1 2.  If  twice 
x  is  equal  to  12,  x  will  be  equal  to  12  divided  by  2,  which 
is  6  ;  therefore,  each  has  6  apples. 

WRITTEX.     * 

Let  x  denote  the  number  of  apples  which  John  has; 
then, 

12 
x  +  x  =  2x  =  12;    hence,    x  =  --  =  6. 

2t 

NOTE. — When  x  is  written  with  the  sign  +  before  it, 
it  is  read  plu-s  x :  and  the  line  above,  is  read,  x  phis  ar 
equals  12. 


10  INTRODUCTION. 

NOTE. — When  x  is  written  by  itself,  it  is  read  one  x» 
and  is  the  same  as,  la; ; 

x  or   he,  means  once             x,  or  one  x, 

2x,  "       twice            x,  or  two  x, 

3x,  "       three  times  x,  or  three  x, 

4x,  "        four  times    x,  or  four  «, 

<fec.,  &c.,  &c. 

2.  What  is  x  +  x  equal  to  ? 

3.  What  is  jc  +  2x   equal  to  ? 

4.  What  is  x  +  2x  +    #   equal  to  ? 

5.  What  is  a;  +  5x  +    x   equal  to  ? 
G.  What  is  x  +  2x  +  3x   equal  to  ? 

7.  James  and  John  together  have  twenty-four  peaches, 
and  one  has  as  many  as  the  other :  how  many  has  each  ? 

ANALYSIS. — Let  x  denote  the  number  which  James  has ; 
then,  since  they  have  an  equal  number,  x  will  also  denote 
the  number  which  John  has,  and  twice  x  will  denote  the 
number  which  both  have,  which  is  24.  If  twice  x  is  equal 
to  24,  x  will  be  equal  to  24  divided  by  2,  which  is  12  ; 
therefore,  each  has  12  peaches. 

•VVEITTKX. 

Let  x  denote  the  number  of  peaches  which  James  has ; 
then, 

24 
x  +  x  =  2x  =  £4 ;    hence,    x  —  —   =  12. 

VERIFICATION. 

A  Verification  is  the  operation  of  proving  that  the  num- 
ber found  will  satisfy  the  conditions  of  the  question.  Thus, 

James1  apples.  John's  apples. 

12  +   12          =   24. 

NOTE. — Let  the  following  questions  be  analyzed,  written, 
and  verified,  in  exactly  the  same  manner  as  the  above. 


M  E  N  T  A  I,       K  X  K  R  0  I  S  K  S  .  11 

8.  William  and  John  together  have  36  pears,  and  one  has 
as  many  as  the  other  :  how  many  has  each  ? 

9.  What  number  added  to  itself  will  make  20  ? 

10.  James  and  John  are  of  the  same  age,  and  the  sum  of 
their  ages  is  32  :  what  is  the  age  of  each? 

11.  Lucy  and  Ann  are  twins,  and  the  sum  of  their  ages 
is  1 6 :  what  is  the  age  of  each  ? 

12.  What  number  is   that  which   added  to  itself  will 
make  30? 

13.  What   number  is  that   which  added  to  itself  will 
make  50? 

14.  Each  of  two  boys  received  an  equal  sum  of  money  at 
Christmas,  and  together  they  received  60  cents :  how  much 
had  each  ? 

15.  What  number  added  to  itself  will  make  100? 

16.  John  has  as  many  pears  as  William;   together  they 
have  72  :  how  many  has  each? 

17.  What  number  added  to  itself  will  give  a  sum  equal 
to  46? 

1 8.  Lucy  and  Ann  have  each  a  rose  bush  with  the  same 
number  of  buds  on  each ;  the  buds  on  both  number  46 : 
how  many  on  each? 


LESSON   H. 

1.  John  and  Charles  together  have  12  apples,  and  Charles 
has  twice  as  many  as  John :  how  many  has  each  ? 

ANALYSIS. — Let  x  denote  the  number  of  apples  which 
John  has ;  then,  since  Charles  has  twice  as  many,  2x  will 
denote  his  share,  and  x  +  2aj,  or  3oj,  will  denote  the 
number  which  they  both  have,  which  is  12.  If  3x  is  equal 
to  12,  x  will  be  equal  to  12  divided  by  3,  which  is  4; 
therefore,  John  has  4  apples,  and  Charles,  having  twice  as 
many,  has  8. 


12  INTKOPUCTION. 

W1UTTEN. 

Let  x  denote  the  number  of  apples  John  has ;  then, 
2x  will  denote  the  number  of  apples  Charles  has ;  and 
x  4-  2a;  =  3x  =  12,    the  number  both  have;  then, 

12 

x  =  —   =     4,    the  number  John  has ;  and, 
3 

2x  =  2  x  4  =  8,   the  number  Charles  has. 

VERIFICATION. 

4  +  8  =   12,    the  number  both  have. 

2.  William  and  John  together  have  48  quills,  and  William 
has  twice  as  many  as  John :  how  many  has  each  ? 

3.  What  number  is  that  which  added  to  twice  itself,  will 
give  a  number  equal  to  60  ? 

4.  Charles'  marbles  added  to  John's  make  3  times  as  many 
as  John  has;  together  they  have  51 :  how  many  has  each? 

ANALYSIS. — Since  Charles'  marbles  added  to  John's  make 
three  times  as  many  as  Charles  has,  Charles  must  have  one 
third,  and  John  two  thirds  of  the  whole. 

Let  x  denote  the  number  which  Charles  has ;  then  2x 
will  denote  the  number  which  John  has,  and  x  +  2#,  or 
So;,  will  denote  what  they  both  have,  which  is  51.  Then,  if 
3x  is  equal  to  51,  x  will  be  equal  to  51  divided  by  3, 
which  is  17.  Therefore,  Charles  has  17  marbles,  and  John, 
having  twice  as  many,  has  34. 

WRITTEN. 

Let   x  denote  the  number  of  Charles'  marbles;  then, 
2x  will  denote  the  number  of  John's  marbles ;  and 

3x  =  51,    the  number  of  both ;  then, 

51 

x  —  —  =  17,    Charles'  marbles;  and 

17  X  2   =  34,    John's  marbles. 


MENTAL      EXERCISES.  13 

5.  What  number  added  to  twice  itself  will  make  75  ? 

6.  "What  number  added  to  twice  itself  will  make  57  ? 

7.  What  number  added  to  twice  itself  will  make  39? 

8.  What  number  added  to  twice  itself  will  give  90  ? 

9.  John  walks  a  certain  distance  on  Tuesday,  twice  aa 
far  on  Wednesday,  and  in  the  two  days  he  walks  27  miles  • 
how  far  did  he  walk  each  day  ? 

10.  Jane's  bush  has  twice  as  many  roses  as  Nancy's:  and 
jn  both  bushes  there  are  36  :  how  many  on  each  ? 

11.  Samuel  and  James  bought  a  ball  for  48  cents  ;  Samuel 
paid  twice  as  much  as  James  :  what  did  each  pay  ? 

12.  Divide  48  into  two  such  parts  that  one  shall  be  double 
the  other. 

13.  Divide  66  into  two  such  parts  that  one  shall  be  double 
the  other. 

14.  The  sum  of  three  equal  numbers  is  12  :  what  are  the 
numbers? 

ANALYSIS.  —  Let  x  denote  one  of  the  numbers;  then, 
since  the  numbers  are  equal,  x  will  also  denote  each  of 
the  others,  and  x  plus  x  plus  #,  or  3x  will  denote  their 
sum,  which  is  12.  Then,  if  3x  is  equal  to  12,  x  will  be 
equal  to  12  divided  by  3,  which  is  4  :  therefore,  the  numbers 
are  4,  4,  and  4. 

WRITTEN. 

Let   x  denote   one  of  the  equal  numbers  ;  then, 
x  +  x  +  x  =  3x  =  12;    and 
12 

" 


VERIFICATION. 

4  +  4  +  4   =   12. 

15.  The  sum  of  three  equal  numbers  is  24  :  what  are  the 
numbers? 

16.  The  sum  of  three  equal  numbers  is  36  :  what  are  the 
numbers  ? 


I  2f  T  14  O  U  U  C  T  I  O  N  . 


17.  The  sum  of  three  equal  numbers  is  54  :  what  .ire  the 
numbers  ? 


LESSON    III. 

1.  What  number  is  that  which  added  to  three  times  itself 
will  make  48  ? 

ANALYSIS. — Let  x  denote  the  number;  then,  3x  will 
denote  three  times  the  number,  and  x  plus  3x,  or  4», 
Avill  denote  the  sum,  which  is  48.  If  4x  is  equal  to  48, 
x  will  be  equal  to  48  divided  by  4,  which  is  12;  there- 
fore, 12  is  the  required  number. 

WRITTEN. 
Let  x  denote  the  number;    then, 

3x  =  three  times  the  number ;    and 
x  +  3x  =  4x  —  48,    the  sum :    then, 

x  —  --  =  12,    the  required  number. 

VERIFICATION. 
12^  34f-12   =   12  +  36   =  48. 

NOTE. — All  similar  questions  are  solved  by  the  same 
form  of  analysis. 

2.  What  number  added  to  4  times  itself  will  give  40  ? 

3.  What  number  added  to  5  times  itself  will  give  42  ? 

4.  What  number  added  to  6  times  itself  will  give  63  ? 

5.  What  number  added  to  7  times  itself  will  give  84  ? 

6.  What  number  added  to  8  times  itself  will  give  81  ? 

7.  What  number  added  to  9  times  itself  will  give  100? 

8.  James  and  John  together  have  24  quills,  and  John  has 
three  times  as  many  as  James :  how  many  has  each  ? 

9.  William  and  Charles  have  64  marbles,  and  Charles  has 
7  times  as  many  as  William :  how  many  has  each  ? 


M  K  N  T  A  L       K  X  K  R  C  J  S  K  S  15 

10.  James  and  John  travel  96  miles,  aud  James  travels 
11  times  as  far  as  John  :  how  far  does  each  travel  ? 

11.  The  sum  of  the  ages  of  a  father  and  son  is  84  years; 
and  the  father  is  3  times  as  old  as  the  son :  what  is  the  age 
of  each  ? 

12.  There  are  two  numbers  of  which  the  greater  is  7 
times  the  less,  and  their  sum  is  72  :  what  are  the  numbers? 

13.  The  sum  of  four  equal  numbers  is  64:  what  are  the 
numbers  ? 

14.  The  sum  of  six  equal  numbers  is  54 :  what  are  the 
numbers  ? 

15.  James  has  24  marbles  ;  he  loses  a  certain  number,  and 
then  gives  away  7  times  as  many  as  he  loses  which  takes  all 
he  has :  how  many  did  he  give  away  ?     Verify. 

16.  William  has  36  cents,  and  divides  them  between  his 
two  brothers,  James  and  Charles,  giving  one,  eight  times  as 
many  as  the  other :  how  many  does  he  give  to  each  ? 

17.  What  is  the  sum  of  x  and   3#?      Of  a;  and   7a;? 
Of  x  and   5x?     Of  x  and   I2x? 


LESSOR   IV. 

1.  If  1  apple  costs  1  cent,  what  will  a  number  of  apples 
denoted  by  x   cost? 

ANALYSIS. — Since  one  apple  costs  1  cent,  and  since  x 
denotes  any  number  of  apples,  the  cost  of  x  apples  will  be 
as  many  cents  as  there  are  apples :  that  is,  x  cents. 

2.  If  1  apple  costs  2  cents,  what  will  x  apples  cost? 

ANALYSIS. — Since  one  apple  costs  2  cents,  and  since  x 
denotes  the  number  of  apples,  the  cost  will  be  twice  as  many 
cents  as  there  are  apples :  that  is  2x  cents. 

3.  If  1  apple  costs  3  cents,  what  will  x  apples  cost  ? 

4.  If  1  lemon  costs  4  cent?,  what  will  x  lemons  cost? 


16  INTRODUCTION. 

5.  If  1  orange  costs  6  cents,  what  will  x  oranges  cost  ? 

6.  Charles  bought  a  certain  number  of  lemons  at  2  cents 
apiece,  and  as  many  oranges  at  3  cents  apiece,  and  paid  in  all 
20  cents :  how  many  did  he  buy  of  each  ? 

ANALYSIS. — Let  x  denote  the  number  of  lemons ;  then, 
since  he  bought  as  many  oranges  as  lemons,  it  will  also 
denote  the  number  of  oranges.  Since  the  lemons  were 
2  cents  apiece,  2x  will  denote  the  cost  of  the  lemons  ;  and 
since  the  oranges  were  3  cents  apiece,  3x  will  denote 
the  cost  of  the  oranges ;  and  2x  +  3.r,  or  5.r,  will  denote 
the  cost  of  both,  which  is  20  cents.  Now,  since  5x  cents 
are  equal  to  20  cents,  x  will  be  equal  to  20  cents  divided  by 
5  cents,  which  is  4 :  hence,  he  bought  4  of  each. 

WRITTEN. 

Let  x  denote  the  number  of  lemons,  or  oranges ;  then, 
2x  =  the  cost  of  the  lemons ;  and 
3x  =  the  cost  of  the  oranges ;  hence, 
2x  +  3x  —  ox  =  20  cents  =  the    cost    of   lemons    and 

oranges ;  hence, 

20  cents  f 

x  =   -  -  —  4,  the  number  of  each. 

5  cents 

VERIFICATION. 

4  lemons  at  2  cents  each,  give,  4x2=     8  cents. 

4  oranges  at  3  cents  each,     "     4  x  3  =±  12  cents. 

Hence,  they  both  cost,  8  cents  +12  cents  =  20  cents. 

7.  A  farmer  bought  a  certain  number  of  sheep  at  4  dollars 
apiece,  and  an  equal  number  of  lambs  at  1  dollar  apiece, 
and  the  whole  cost  60  dollars:  how  many  did  he  buy  of 
each  ? 

8.  Charles  bought  a  certain  number  of  apples  at  1  cent 
apiece,  and  an  equal  number  of  oranges  at  4  cents  apiece,  and 
paid  60  cents  in  all :  how  many  did  he  buy  of  each  ? 


M  K  X  T  A  I.       E  X  E  R  C  I  S  t  S  .  17 

9.  James  bought  an  equal  number  of  apples,  pears,  and 
lemons ;  he.  paid  1  cent  apiece  fqr  the  apples,  2  cents  apiece 
for  the  pears,  and  3  cents  apiece  for  the  lemons,  and  paid 
72  cents  in  all :  how  many  did  he  buy  of  each  ?     Verify. 

10.  A  farmer  bought  an  equal  number  of  sheep,  hogs, 
and  calves,  for  which  he  paid  108  dollars;  he  paid  3  dollars 
apiece   for  the  sheep,  5  dollars  apiece  for  the  hogs,  and 
4  dollars  apiece  for  the  calves :  how  many  did  he  buy  of 
each? 

11.  A  farmer  sold   an   equal   number   of  ducks,  geese, 
and  turkeys,  for  which  he  received  90  shillings.    The  ducks 
brought  him  3  shillings  apiece,  the  geese  5,  and  the  turkeys 
7 :  how  many  did  he  sell  of  each  sort  ? 

12.  A  tailor  bought,  for  one  hundred  dollars,  two  pieces 
of  cloth,  each  of  which  contained  an  equal  number  of  yards. 
For  one  piece  he  paid  3  dollars  a  yard,  and  for  the  other 
2  dollars  a  yard :  how^many  yards  in  each  piece  ? 

13.  The  sum  of  three  numbers  is  28  ;  the  second  is  twice 
the  first,  and  the  third  twice  the  second:    what  are  the 
numbers  ?     Verify. 

14.  The  sum  of  three  numbers  is  64  ;  the  second  is  3  times 
the  first,  and  the  third  4  times  the  second :  what  are  the 
numbers  ? 


LESSON    V. 

1.  If  1  yard  of  cloth  costs  x  dollars,  what  will  2  yards 
cost  ? 

ANALYSIS. — Two  yards  of  cloth  will  cost  twice  as  much  as 
one  yard.  Therefore,  if  1  yard  of  cloth  costs  x  dollars, 
2  yards  will  cost  twice  x  dollars,  or  2x  dollars. 

2.  If  1  yard  of  cloth  costs  x  dollars,  what  will  3  yards 
cost  ?     Why  ? 


18  I  N  T  K  O  I)  U  C  T  I  O  N  . 

3.  If  1  orange  costs  x  cents,  what  wil)    i   crai>g6b  cost? 
Why  ?     8  oranges  ? 

4.  Charles  bought  3  lemons  and  4  ora.ige?,  for  which  he 
paid  22  cents.     He  paid  twice  as  much  for  an  orange  as  for 
a  lemon  :  what  was  the  price  of  each  ? 

ANALYSIS. — Let  x  denote  the  price  of  a  lemon ;  then,  2x 
will  denote  the  price  of  an  orange ;  3x  will  denote  the  cost, 
of  3  lemons,  and  8x  the  cost  of  4  oranges ;  hence,  3x  plus 
8z,  or  lla;,  will  denote  the  cost  of  the  lemons  and  oranges, 
which  is  22  cents.  If  lla:  is  equal  to  22  cents,  x  is  equal  to 
22  cents  divided  by  11,  which  is  2  cents:  therefore,  the 
price  of  1  lemon  is  2  cents,  and  that  of  1  orange  4  cents. 

WRITTEN". 

Let  x  denote  the  price  of  1  lemon ;  then, 
2x  =  1  orange ;  and, 

3o;  -f  8x  ==   llx  =  22  cts.,  the  cost  of  lemons  and  oranges; 

22  cts 
hence,  x  =  — — — -  =  2  cts.,  the  price  of  1  lemon ; 

and,  2x2  =  4  cts.,  the  price  of  1  orange. 

VERIFICATION. 

3x2=     6  cents,  cost  of  lemons, 
4  x  4  =   16  cents,  cost  of  oranges. 
22  cents,  total  cost. 

5.  James  bought  8  apples  and  3  oranges,  for  which  he 
paid  20  cents.     He  paid  as  much  for  1  orange  as  for  4  apples: 
what  did  he  pay  for  one  of  each  ? 

6.  A  farmer  bought  3  calves  and  7  pigs,  for  which  he  paid 
1 9  dollars.     He  paid  four  times  as  much  for  a  calf  as  for  a 
pig :  what  was  the  price  of  each  ? 

7.  James  bought  an  apple,  a  peach,  and  a  pear,  for  which 
he  paid  6  cents.     He  paid  twice  as  much  for  the  peach  as  for 


M  ]•:  N  T  A  L      EXERCISES.  19 

the  apple,  and  three  times  as  much  for  the  pear  as  for  the 
apple :  what  was  the  price  of  each  ? 

8.  William  bought  an  apple,  a  lemon,  and  an  orange,  for 
which  he  paid  24  cents.     He  paid  twice  as  much  for  the 
lemon  as  for  the  apple,  and  3  times  as  much  for  the  orange 
as  for  the  apple :  what  was  the  price  of  each  ? 

9.  A  farmer  sold  4  calves  and  5  cows,  for  which  he  received 
120  dollars.     He  received  as  much  for  1  cow  as  for  4  calves : 
what  was  the  price  of  each  ? 

10.  Lucy  bought  3  pears  and  5  oranges,  for  which  she 
paid  26  cents,  giving  twice  as  much  for  each  orange  as  for 
each  pear:  what  was  the  price  of  each? 

11.  Ann  bought  2  skeins  of  silk,  3  pieces  of  tape,  and  a 
penknife,  for  which  she  paid  80  cents.     She  paid  the  same 
for  the  silk  as  for  the  tape,  and  as  much  for  the  penknife  as 
for  both  :  what  was  the  cost  of  each  ? 

12.  James,  John,  and   Charles  are   to   divide   56  cents 
among  them,  so  that  John  shall  have  twice  as  many  as 
James,  and  Charles  twice  as  many  as  John:  what  is  the 
share  of  each  ? 

13.  Put  54  apples  into  three  baskets,  so  that  the  second 
shall  contain  twice  as  many  as  the  first,  and  the  third  as 
many  as  the  first  and  second :  how  many  will  there  be  in 
each. 

14.  Divide  60  into  four  such  parts  that  the  second  shall 
be  double  the  first,  the  third  double  the  second,  and  the 
fourth  double  the  third  :  what  are  the  numbers  ? 


LESSON    VL 

1.  If   2x  +  x    is  equal  to    3cc,    what  is    Sx  —  x    equal 
to  ?     Written,  3x  —  x  —   2x. 

2.  What  is    4x  —  x    equal  to  ?     Written, 

4x  —  x  =  3x. 


20  INTRODUCTION. 

3.  What  is  Sx  minus  Qx  equal  to  ?     Written, 

8x  —  6x  =  2x. 

4.  What  is    12a;  —  9x    equal  to?  Ans.  3% 

5.  What  is    15x  —  Ix    equal  to  ? 

6.  What  is    11  x  —  ISx    equal  to?  Ans.  4a\ 

7.  Two  men,  who  are  30  miles  apart,  travel  towards  each 
other  ;  one  at  the  rate  of  2  miles  an  hour,  and  the  other  at 
the  rate  of  3  miles  an  hour :  how  long  before  they  will  meet? 

ANALYSIS. — Let  x  denote  the  number  of  hours.  Then, 
since  the  time,  multiplied  by  the  rate,  will  give  the  distance, 
2x  will  denote  the  distance  traveled  by  the  first,  and  3x 
the  distance  traveled  by  the  second.  But  the  sum  of  the 
distances  is  30  miles ;  hence, 

2x  +  3x  =  5x  =  30  miles ; 

and  if  5x  is  equal  to  30,  a;  is  equal  to  30  divided  by  6, 
which  is  6 :  hence,  they  will  meet  in  6  hours. 

WRITTEN". 

Let  x  denote  the  time  in  hours ;  then, 

2*  =  the  distance  traveled  by  the  1st;  and 
3*  =  "  "  2d. 

By  the  conditions, 

2a  +  3x  =  5x  =  30  miles,  the  distance  apart ; 

30 

hence,  x  =  —  =  6  hours. 

5 

VERIFICATION. 

2x6  =  12  miles,  distance  traveled  by  the  first. 
3x6  —  18  miles,  distance  traveled  by  the  second. 
30  miles,  whole  distance. 

8.  Two  persons  are  10  miles  apart,  and  are  traveling  ID 
the  same  direction ;  the  first  at  the  rate  of  3  miles  an  hour, 
and  the  second  at  the  rate  of  5  miles :  how  long,  before  the 
second  will  overtake  the  first  ? 


M  K  N  T  A  L       ]•;  X  K  U  C  I  8  E  8  .  21 

ANALYSIS. — Let  x  denote  the  time,  in  hours.  Then,  Sx 
will  denote  the  distance  traveled  by  the  first  in  x  hours; 
and  5x  the  distance  traveled  by  the  second.  But  Avhen 
the  second  overtakes  the  first,  he  will  have  traveled  10  miles 
more  than  the  first :  hence, 

5x  —  3x  —  2x  =   10 ; 

if  2x  is  equal  to  10,  x  is  equal  to  5  :  hence,  the  second  will 
overtake  the  first  in  5  hours. 

W1UTTEN. 

Let  x  denote  the  time,  in  hours :  then, 

3x  =  the  distance  traveled  by  the  1st; 
and,    5x  =  "  "  2d; 

and,    5x  —  3x  =  2x  =  10  hours; 

10 
or,  x  =  -  -  =     5  hours. 

2i 

VERIFICATION. 

3x5   =   15  miles,  distance  traveled  by  1st. 
5  X  5   =  25  miles,  "•  "  2d. 

25  —  15  =  10  miles,  distance  apart. 

9.  A   cistern,  holding  100  hogsheads,  is  filled  by  two 
pipes ;  one  discharges  8  hogsheads  a  minute,  and  the  other 
12  :  in  what  time  will  they  fill  the  cistern  ? 

10.  A  cistern,  holding  120  hogsheads,  is  filled  by  3  pipes ; 
the  first  discharges  4  hogsheads  in  a  minute,  the  second  7, 
and  the  third  1 :  in  what  time  will  they  fill  the  cistern  ? 

11.  A  cistern  which  holds  90  hogsheads,  is  filled  by  a  pipe 
which  discharges  10  hogsheads  a  minute;   but  there  is  a 
waste  pipe  which  loses  4  hogsheads  a  minute :  how  long 
will  it  take  to  fill  the  cistern  ?  m 

12.  Two  pieces  of  cloth  contain  each  an  equal  number  of 
yards ;  the  first  cost  3  dollars  a  yard,  and  the  second  5,  and 
both  pieces  cost  96  dollars  :  how  many  yards  in  each? 

13.  Two  pieces  of  cloth  contain  each  an  equal  number  of 
yards ;  the  first  cost  7  dollars  a  yard,  and  the  second  5 ;  the  first 


22  IN  T  R  O  D  C  C  T  I  O  X  . 

cost  CO  dollars  more  than  the  second :  how  many  yards  in 
each  piece  ? 

14.  John  bought  an  equal  number  of  oranges  and  lemons 
the  oranges  cost  him  5  cents  apiece,  and  the  lemons  3 ;  and 
he  paid  56  cents  for  the  whole:  how  many  did  he  buy  of 
each  kind  ? 

15.  Charles  bought  an  equal  number  of  oranges   and 
lemons;    the   oranges   cost  him  5  cents   apiece,   and    the 
lemons  3  ;  he  paid  14  cents  more  for  the  oranges  than  for 
the  lemons :  how  many  did  he  buy  of  each  ? 

16.  Two  men  work  the  sa.me  number  of  days,  the  one 
receives  1  dollar  a  day,  and  the  other  two :  at  the  end  of 
the  time  they  receive  54  dollars :  how  long  did  they  work  ? 


LESSON    VIE. 

1.  John  and  Charles  together  have  25  cents,  and  Charles 
has  5  more  than  John :  how  many  has  each  ? 

ANALYSIS. — Let  x  denote  the  number  which  John  has ; 
then,  x  +  5  will  denote  the  number  which  Charles  has,  and 
x  -f-  x  +  5,  or  2x  +  5,  will  be  equal  to  25,  the  number 
they  both  have.  Since  2x  -f  5  equals  25,  2x  will  be 
equal  to  25  minus  5,  or  20,  and  x  will  be  equal  to  20 
divided  by  2,  or  10:  therefore,  John  has  10  cents,  and 
Charles  15. 

WRITTEN-. 

Let  x  denote  the  number  of  John's  cents ;  then, 
x  4-  5  =  "  Charles'  cents ;  and, 

*  x  +  x  +  5  =  25,  the  number  they  both  have ;  or, 
2x  +  5   =  25  ;     and, 

2x  =  25  —  5   =  20 ;     hence, 

20 
x  =  —  =10,  John's  number;  and, 

I 

10  +  5  z=  15,  Charles'  number. 


MENTAL      EXERCISES.  23 


VERIFICATION. 
John1!.  Charles1. 

10  +15  =  25,    the  sum. 

Charles'.  John's. 

15  -   10          =     '5,    the  difference. 

2.  James  and  John  have  30  marbles,  and  John  has  4  more 
tnan  James  :  how  many  has  each  ? 

3.  William  bought  60  oranges  and  lemons ;    there  were 
20  more  lemons  than  oranges :    how  many  were  there  of 
each  sort  ? 

4.  A  farmer  has  20  more  cows  than  calves;  in  all  he  has 
36  :  how  many  of  each  sort  ? 

5.  Lucy  has  28  pieces  of  money  in  her  purse,  composed 
of  cents  and  dimes ;  the  cents  exceed  the  dimes  in  number 
by  16  :  how  many  are  there  of  each  sort  ? 

6.  What  number  added  to  itself,  and  to  9,  will  make  29  ? 

7.  What  number  added  to  twice  itself,  and  to  4,  will 
make  25  ? 

8.  What  number  added  to  three  times  itself,  and  to  12, 
will  make  60  ? 

9.  John  has  five  times  as  many  marbles  as  Charles,  and 
what  they  both  have,  added  to  14,  makes  44  :  how  many  has 
each  ? 

10.  There  are  three  numbers,  of  which  the  second  is  twice 
the  first,  and  the  third  twice  the  second,  and  when  9  is 
added  to  the  sum,  the  result  is  30 :  what  are  the  numbers? 

11.  Divide  13  into  two  such  parts  that  the  second  shall 
be  two  more  than  double  the  first :  what  are  the  parts  ? 

12.  Divide  50  into  three  such  parts  that  the  second  shall 
be  tAvice  the  first,  and  the  third  exceed  six  times  the  first 
by  4  :  what  are  the  parts? 

13.  Charles  has  twice  as  many  cents  as  James,  and  John 


24  I  N  T  li  O  D  U  C  T  I  O  N  . 

has  twice  as  many  as  Charles ;  if  7  be  added  to  what  they 
all  have,  the  sum  will  be  28  :  how  many  has  each  ? 

14.  Divide  15  into  three  such  parts  that  the  second  shall 
be  3  times  the  first,  the  third  twice  the  second,  and  5  over : 
what  are  the  numbers  ? 

15.  An  orchard  contains  three  kinds  of  trees,  apples,  pears, 
and  cherries;  there  are  4  times  as  many  pears  as  apples, 
twice  as  many  cherries  as  pears,  and  if  14  be  added,  the 
number  will  be  40 ;  how  many  are  there  of  each  ? 


LESSON    VIII. 

1.  John  after  giving  away  5  marbles,  had  12  left:  how 
many  had  he  at  first  ? 

ANALYSIS. — Let  x  denote  the  number ;  then,  x  minus  5 
will  denote  what  he  had  left,  which  was  equal  to  12.  Since 
x  diminished  by  5  is  equal  to  12,  x  will  be  equal  to  12, 
increased  by  5 ;  that  is,  to  1 7 :  therefore,  he  had  1 7  marbles. 

WRITTEN. 

Let  x  denote  the  number  he  had  at  first;  then, 
x  —  5   =  12,    what  he  had  left;  and 

x  =  12  +  5  =  17,    what  he  first  had. 

VERIFICATION. 

17  —  5  =  12,    what  were  left. 

2.  Charles  lost  6  marbles  and  has  9  left :  how  many  had 
he  at  first  ? 

3.  William  gave  15  cents  to  John,  and  had  9  left :  how 
many  had  he  at  first  ? 

4.  Ann  plucked  8  buds  from  her  rose  bush,  and  there 
-?rere  1 9  left :  how  many  were  there  at  first  ? 


M  E  X  T  A  L       EXERCISES.  25 

5.  William  took  27  cents  from  his  purse,  and  there  \verr 
1 3  left :  how  many  were  there  at  first  ? 

6.  The  sum  of  two  numbers  is  14,  and  their  difference  is  2: 
what  are  the  numbers? 

ANALYSIS. — The  difference  of  two  numbers,  added  to  the 
less,  will  give  the  greater.  Let  x  denote  the  less  number ; 
then,  x  +  2,  will  denote  the  greater,  and  x  +  x  +  2, 
will  denote  their  sum,  Avhich  is  14.  Then,  2x  +  2  equals 
14;  and  2x  equals  14  minus  2,  or  12:  hence,  x  equals 
12  divided  by  2,  or  6 :  hence,  the  numbers  are  6  and  8. 

VERIFICATION. 

6  +  8   =   14,  their  sum  ;  and 
8  —  6   =     2,  their  difference. 

7.  The  sum  of  two  numbers  is  18,  and  their  difference  0  :, 
what  are  the  numbers  ? 

8.  James  and  John  have  26  marbles,  and  James  has  4  more 
than  John  :  how  many  has  each  ? 

9.  Jane  and  Lucy  have  16  books,  and  Lxicy  has  8  more 
than  Jane  :  how  many  has  each  ? 

10.  William   bought  an  equal   number  of  oranges   and 
lemons ;  Charles  took  5  lemons,  after  which  William  had  but 
25  of  both  sorts  :  how  many  did  he  buy  of  each  ? 

11.  Mary  has  an  equal  number  of  roses  on  each  of  two 
bushes ;  if  she  takes  4  from  one  bush,  there  will  remain  24 
on  both  :  how  many  on  each  at  first  ? 

12.  The  sum  of  two  numbers  is  20,  and  their  difference 
is  6  :  what  are  the  numbers  ? 

ANALYSIS. — If  x  denotes  the  greater  number,  x  —  6  will 
denote  the  less,  and  x  +  x  —  6  will  be  equal  to  20;  hence, 
2x   equals   20  +  6,    or  26,  and  x   equals  26  divided  by  2, 
equals  13  ;  hence  the  numbers  are  13  and  7. 
2 


I  N  T  It  O  U  U  O  T  I  O  N  . 


Let  x  denote  the  greater  ;  then, 

x  —  6   =  the  less  ;  and 
x  -j-  x  —  6  =  20,  their  sum  ;  hence, 

2x  =   20  +  6   =   26  ;  or, 

26 
x  =  —  —   13  ;  and  13  —  6   =    7. 

VERIFICATION. 

13  +  7   =  20;  and,  13  —  7  =  6. 

13.  The  sum  of  the  ages  of  a  father  and  son  is  60  yeais, 
and  their  difference  is  just  half  that  number  :  what  are  theij 
ages? 

14.  The  sum  of  two  numbers  is  23,  and  the  larger  lacks 
1  of  being  7  times  the  smaller  :  what  are  the  numbers  ? 

15.  The  sum  of  two  numbers  is  50  ;  the  larger  is  equal  to 
10  times  the  less,  minus  5  :  what  are  the  numbers  ? 

16.  John  has  a  certain  number  of  oranges,  and  Charles 
has  four  times  as  many,  less  seven  ;  together  they  have  53  : 
how  many  has  each  ? 

17.  An  orchard  contains  a  certain  number  of  apple  trees, 
and  three  times  as  many  cherry  trees,  less  6  ;  the  whole  num- 
ber is  30  :  how  many  of  each  sort? 


LESSON  IX. 

1.  If  x  denotes  any  number,  and  1  be  added  to  it,  what 
will  denote  the  sum  ?  Ans.  x  -f  1 . 

2.  If  2  be  added  to  x,  what  will  denote  the  sum  ?    If  3 
be  added,  what  ?    If  4  be  added  ?  <fec. 

If  to  John's  marbles,  one  marble  be  added,  twice  his  num. 
ber  will  be  equal  to  10  :  how  many  had  he  ? 

ANALYSIS. — Let  x  denote  the  number ;  then,  x  -\-  1  will 
denote  the  number  after  1  is  added,  and  twice  this  number, 


M  K  N  T  A  L       EXERCISKS.  27 

or   2x  +  2,  will  be  equal  to  10.     If  2x  +  2  is  equal  to  10, 
2x  will  be  equal  to  10  minus  2,  or  8 ;  or  x  will  be  equal  to  4. 

WRITTEN; 
Let  x  denote  the  number  of  John's  marbles ;  then, 

x  -f  1   =  the  number,  after  1  is  added ;  and 
2(oj  +  1)   =  2a?  +  2  =  10;  hence, 

8 
2x  =  10  —  2  ;  or  x  =  -   =  4. 

2i 

VERIFICATION. 

2(4  +  1)   =   2  X  5   =  10. 

4.  Write  x  -f  2  multiplied  by  3.  Ans.  3(x  +  2). 
What  is  the  product  ?  Ans.  3a;+6. 

5.  Write  x  +  4  multiplied  by  5.  Ans.  5(x  +  4). 
What  is  the  product  ?  Ans.  5x  +  20. 

6.  Write  x  -f-  3  multiplied  by  4.  ,4ws.  4(ce  -f  3). 
What  is  the  product?  Ans.  4x  -+-  12. 

7.  Lucy  has  a  certain  number  of  books ;  her  father  gives 
her  two  more,  when  twice  her  number  is  equal  to  14  :  how 
many  has  she  ? 

8.  Jane  has  a  certain  number  of  roses  in  blossom ;  two 
more  bloom,  and  then  3  tunes  the  number  is  eqiial  to  15  : 
how  many  were  in  blossom  at  first  ? 

9.  Jane  has  a  certain  number  of  handkerchiefs,  and  buys 
4  more,  when  5  times  her  number  is  equal  to  45  :  hoAV  many 
had  she  at  first  ? 

10.  John  has  1   apple  more  than  Charles,  and  3  times 
John's,  added  to  what  Charles  has,  make  15  :  how  many 
has  each  ? 

ANALYSIS. — Let  x  denote  Charles'  apples ;  then  x  +  1  will 
denote  John's ;  and  x  +  1  multiplied  by  3,  added  to  ce,  or 
!te  +  3  +  »,  will  be  equal  to  15,  what  they  both  had ;  hence, 
4«  +  3  equals  15;  and  4a;  equals  15  minus  3,  or  12;  and 
gr.  =  4.  Write,  and  verify. 


28  I  N  T  K  O  1)  U  C  T  ION. 

]  1.  James  has  two  marbles  more  than  William,  and  twice 
his  marbles  plus  twice  William's  are  equal  to  16  :  how  many 
has  each  ? 

12.  Divide  20  into  two  such  parts  that  one  part  shall  ex- 
ceed the  other  by  4. 

13.  A  fruit-basket  contains  apples,  pears,  and  peaches; 
there  are   2  more  pears  than  apples,  and  twice  as  many 
peaches  as  pears;  there  are  22  in  all:  how  many  of  each 
sort  ? 

14.  What  is  the  sum  of  x  +  3x  +  2(x  +  1)  ? 

15.  What  is  the  snm  of  2 (x  +  1)  +  l(x  +  1)  +  x? 

16.  What  is  the  sum  of  x  +  5(x  +  8)  ? 

17.  The  sum  of  two  numbers  is  11,  and  the  second  is  equal 
to  twice  the  first  plus  4  :  what  are  the  numbers  ? 

18.  John  bought  3  apples,  3  lemons,  and  3  oranges,  for 
which  he  paid  27  cents ;  he  paid  1  cent  more  for  a  lemon 
than  for  an  apple,  and  1  cent  more  for  an  oi'ange  than  for  a 
lemon  :  what  did  he  pay  for  each  ? 

19.  Lucy,  Mary,  and  Ann,  have  15  cents;  Mary  has  1 
more  than  Lucy,  and  Ann  twice  as  many  as  Mary  ? 


LESSON  X. 

-  1.  If  x  denote  any  number,  and  1  be  subtracted  from  it, 
what  will  denote  the  difference?  Ans.  x  —  1. 

If  2  be  subtracted,  what  Avill  denote  the  difference  ?  If 
3  be  subtracted  ?  4  ?  &c. 

2.  John  has  a  certain  number  of  marbles ;  if  1  be  taken 
away,  twice  the  remainder  will  be  equal  to  12:  how  many 
has  he  ? 

ANALYSIS. — Let  x  denote  the  number ;  then,  x  —  1  wilJ 
denote  the  number  after  1  is  taken  away ;  and  twice  this 
number,  or  2(ar .  —  1)  —  2a-  —  2,  will  be  equal  to  12.  If  2x 


M  K  N  T  A  -L       K  X  K  H  O  I  S  1C  S  .  29 

diminished  by  2  is  equal  to  12,  2x  is  equal  to  12  plus  2,  or 
14  ;  hence,   x  equals  14  divided  by  2,  or  7. 

WRITTEN. 

Let  x   denote  the  number ;  then, 

x  —  1   =  the  number  which  remained,  and 
2(x  —  1)   =  2x  —  2  =   12  ;  hence, 

14 

2x  =   12  4-  2,  or  14  :  and  x  =  —  =  7. 

2 

VERIFICATION. 

2(7-1)   =   14  -  2  =  12  ;  also,  2(7  —  1)   =  2  X  6  =  12 

3.  Write  3  times  x  —  1.  Ans.  3(x  —  1). 
What  is  the  product  equal  to  ?  Ans.  3x  —  3. 

4.  Write  4  times  x  —  2.  Ans.  4(x  —  2). 
What  is  the  product  equal  to  ?  Ans.  Ix  —  8. 

5.  Write  5  times  x  —  5.  Ans.  5(x  —  5). 
What  is  the  product  equal  to  ?  Ans.  5x  —  25. 

6.  If  x  denotes  a  certain  number,  will  x    —  1    denote  a 
greater  or  less  number  ?  how  much  less  ? 

7.  If  x  —  1  is  equal  to  4,  Avhat  will  x  be  equal  to  ? 

Ans.  4  +  1,  or  5. 

8.  If  x  —  2  is  equal  to  6,  what  is  x  equal  to  ? 

9.  James  and  John  together  have  20  oranges ;  John  has 
6  less  than  James:  how  many  has  each? 

10.  A  grocer  sold  12  pounds  of  tea  and  coffee ;  if  the  tea 
be  diminished  by  3  pounds,  and  the  remainder  multiplied  by 
2,  the  product  is  the  number  of  pounds  of  coffee :  how  many 
pounds  of  each  ? 

11.  Ann  has  a  certain  number  of  oranges;  Jane  has  1  less, 
and  twice  her  number  added  to  Ann's  make  13  :  how  many 
has  each  ? 

AXALTSIS. — Let  x  denote  the  number  of  oranges  which 
Ana  has;  then,  x  —  1  will  denote  the  number  Jane  has, 


30  INTRODUCTION. 

and  x  +  2x  —  2,  or  3x  —  2,  will  denote  the  number  both 
have,  which  is  13.  If  3x  —  2  equals  13,  3x  will  be  equal 
to  13  +  2,  or  15  ;  and  if  3x  is  equal  to  15,  x  will  be  equal 
to  15  divided  by  3,  which  is  5  :  hence,  Ann  has  5  oranges 
and  Jane  4. 

WRITTEN. 

Let   x   denote  the  number  Ann  has  ;  then, 
x  —  I   =  the  number  Jane  has;  and 
2(x  —  1)   =  2x  —  2  =  tAvice  what  Jane  has;  also, 
*  +  2x  —  2  =  3x  —  2   =  13;  hence, 

15 

3x  =    13  +  2   =   15;  or  x  =  —   =   5. 

3 

VERIFICATION. 

5—4   =   1  ;  and  2x4  +  5   =   13. 

12.  Charles  and  John  have  20  cents,  and  John  has  6  less 
thaii  Charles :  how  many  has  each  ? 

13.  James  has  twice  as  many  oranges  as  lemons  in  his  bas- 
ket, and  if  5  be  taken  from  the  whole  number,  19  will  re- 
main :  how  many  had  he  of  each  ? 

14.  A  basket  contains  apples,  peaches,  and  pears;  29  in 
all.     If  1  be  taken  from  the  number  of  apples,  the  remainder 
will  denote  the  number  of  peaches,  and  twice  that  remainder 
will  denote  the  number  of  pears :  how  many  are  there  of 
each  sort  ? 

15.  If  2x  —  5  equals  15,  what  is  the  value  of  x? 

16.  If  4x  —  5  is  equal  to  11,  what  is  the  value  of  a;? 

17.  If  5x  —  12  is  equal  to  18,  what  is  the  value  of  x? 

18.  The  sum  of  two  numbers  is  32,  and  the  greater  ex- 
ceeds the  less  by  8  :  what  are  the  numbers  ? 

19.  The  sum  of  2  numbers  is  9  ;  if  the  greater  number 
be  diminished  by  5,  and  the  remainder  multiplied  by  3,  the 
product  will  be  the  less  number  :  what  are  the  numbers? 

20.  There  are  three  numbers  such  that  1  taken  from  the 


MENTAL      EXERCISES.  31 

first  will  give  the  second ;  the  second  multiplied  by  3  will 
give  the  third ;  and  their  sum  is  equal  to  26 :  what  are  the 
numbers  ? 

21.  John  and  Charles  together  have  just  31  oranges;  if 
I  be  taken  from  John's,  and  the  remainder  be  multiplied  by 
5,  the  product  will  be  equal  to  Charles'  number:  how  many 
has  each  ? 

22.  A  basket  is  filled  with  apples,  lemons,  and  oranges,  in 
all  26 ;  the  number  of  lemons  exceed  the  apples  by  2,  and 
the  number  of  oranges  is  double  that  of  the  lemons :  how 
many  are  there  of  each  ? 


LESSON    XL 

1.  John  has  a  certain  number  of  apples,  the  half  of  which 
is  equal  to  1 0 :  how  many  has  he  ? 

ANALYSIS. — Let  x  denote  the  number  of  apples ;  then, 
x  divided  by  2  is  equal  to  10 ;  if  one  half  of  x  is  equal  to 
1 0,  twice  one-half  of  cc,  or  #,  is  equal  to  twice  10,  which  is 
20 ;  hence,  x  is  equal  to  20. 

NOTE. — A  similar  analysis  is  applicable  to  any  one  of  the 
fractional  units.  Let  each  question  be  solved  according  to 
the  analysis. 

2.  John  has  a  certain  number  of  oranges,  and  one-third  of 
his  number  is  15  :  how  many  has  he? 

3.  If  one-fifth  of  a  number  is  C,  what  is  the  number? 

4.  If  one-twelfth  of  a  number  is  9,  what  is  the  number? 
5^  What  number  added  to  one-half  of  itself  will  give  a 

sum  equal  to  12? 

ANALYSIS. — Denote  the  number  by  x ;  then,  x  plus  one- 
half  of  a;  equals  12.  But  x  plus  one-half  of  a;  equals  three 
halves  of  x:  hence,  three  halves  of  x  equal  12.  If  three 
halves  of  a;  equal  12,  one-half  of  x  equals  one-third  of  12, 


32  INTRODUCTION 

or  4.     If  one-half  of  x    equals  4,  x  equals  twice  4,  or  8  ' 
hence,  x  equals  8. 

WRITTEN. 

Let  x  denote  the  number;  then, 

1  3 

x  +  -x  =   -x  =   12;    then, 

-x  —     4,    or    x  =  8. 

VERIFICATION. 

8  4-  I  =  8  +  4  =  12. 

6.  What  number  added  to  one-third  of  itself  will  give  a 
sum  equal  to  12? 

'    7.  What  number  added  to  one-fourth  of  itself  will  give 
a  sum  equal  to  20  ? 

8.  What  number  added  to  a  fifth  of  itself  will  make  24  ? 

9.  What  number   diminished  by  one-half  of  itself  will 
leave  4  ?    Why  ? 

10.  What  number  diminished  by  one-third  of  itself  will 
leave  6  ? 

11.  James  gave  one-seventh  of  his  marbles  to  William, 
and  then  has  24  left :  how  many  had  he  at  first  ? 

12.  What  number  added  to  two-thirds  of  itself  will  give 
a  sum  equal  to  20  ? 

13.  What  number  diminished  by  three-fourths  of  itself 
will  leave  9  ? 

14.  What  number   added  to   five-sevenths  of  itself  will 
make  24  ? 

15.  What  number  diminished  by  seven-eighths  of  itself 
will  leave  4  ? 

16.  What   number   added   to   eight-ninths  of  itself  wiJJ 
make  34  ? 


ELEMENTARY   ALGEBRA. 


CHAPTER  I, 

DEFINITIONS   AND   EXPLANATORY   SIGNS. 

1.  QUANTITY  is  anything  that  can  be  measured,  as  num- 
ber, distance,  weight,  time,  &c. 

To  measure  a  thing,  is  to  find  how  many  times  it  contains 
some  other  thing  of  the  same  kind,  taken  as  a  standard.  The 
assumed  standard  is  called  the  unit  of  measure. 

2.  MATHEMATICS  is  the  science  which  treats  of  the  pro- 
perties and  relations  of  quantities. 

In  pure  mathematics,  there  are  but  eight  kinds  of  quantity, 
and  consequently  but  eight  kinds  of  UNITS,  viz. :  Units  of 
Number ;  Units  of  Currency  ;  Units  of  Length  ;  Units  of 
Surf  act ;  Units  of  Volume;  Units  of  Weight;  Units  of 
Time  ;  and  Units  of  Angular  Measure. 

8.  ALGEHRA  is  a  branch  of  Mathematics  in  which  the 
quantities  considered  are  represented  by  letters,  and  the 
operations  to  be  performed  are  indicated  by  signs. 

1.  What  is  quantity?     What  is  the  operation  of  measuring  a  thing ? 
What  is  the  assumed  standard  called  ? 

2.  What  is  Mathematics  ?     How  many  kinds  of  quantity  are  there  iu 
the  pure  mathematics?     Name  the  units  of  those  quantities. 

:5.  What  is  Algebra? 
1* 


34:  ELK  M  E  X  T  A  li  Y       A  L  G  K  B  H  A  . 

4.  The  quantities  employed  in  Algebra  are  of  two  kinds, 
Known  and  Unknown: 

Known  Quantities  are  those  whose  values  are  given ; 

and 

Unknown  Quantities  are  those  whose  values  are  re- 
quired. 

Known  Quantities  are  generally  represented  by  the  lead- 
ing letters  of  the  alphabet,  as,  a,  #,  c,  &c. 

Unknown  Quantities  are  generally  represented  by  the. 
final  letters  of  the  alphabet ;  as,  x,  y,  z,  &c. 

When  an  unknown  quantity  becomes  known,  it  is  often 
denoted  by  the  same  letter  with  one  or  more  accents ;  as, 
ce',  a;",  x".  These  symbols  are  read:  x  prime  ;  x  second; 
x  third,  &c. 

5.  The  SIGN  OF  ADDITION,   +,  is  called  plus.     When 
placed  between  two  quantities,  it  indicates  that  the  second 
is  to  be  added  to  the  first.     Thus,  a  +  5,  is  read,  a  plus  #, 
and  indicates  that  b  is  to  be  added  to  a.     If  no  sign  is 
written,  the  sign  +  is  understood. 

The  sign  +,  is  sometimes  called  \kepositive  sign,  and  the 
quantities  before  which  it  is  written  are  called  2)ositiee  quan- 
tities, or  additive  quantities. 

6.  The  SIGN  OF  SUBTRACTION,  — ,  is  called  minus.   When 
placed  between  two  quantities,  it  indicates  that  the  second 
is  to  be  subtracted  from  the  first.     Thus,  the  expression, 

4.  How  many  kinds  of  quantities  arc  employed  in  Algebra  ?     How  are 
they  distinguished  ?     What  are  known  quantities  ?     What  are  unknown 
quantities?     By  what  are  the  known  quantities  represented?     By  what 
are  the  unknown  quantities  represented?     When  an  unknown  quantity 
becomes  known,  how  Is  it  often  denoted? 

5.  What  is  the  sign  of  ailditiou  called?     When  placed  between  two 
quantities,  what  does  it  indicate  ? 

G.  What  is  the  sign  of  subtraction  called  ?  When  placed  between  two 
quantities,  what  does  it  indicate  ? 


DEFINITION       OK      T  K  K  M  S .  36 

c  —  d,  rcxd  c  minus  d,  indicates  that  d  is  to  be  subtracted 
from  c.  ff  a  stands  for  6,  and  d  for  4,  then  a  —  d  is  equal 
to  6  —  4,  which  is  equal  to  2. 

The  v.gn  — ,  is  sometimes  called  the  negative  sign,  and  the 
quantities  before  which  it  is  written  are  called  negative  quan- 
tities, or  subtractive  quantities. 

7.  The  SIGN  OF  MULTIPLICATION,  x ,  is  read,  multiplied 
by,  or  into.  When  placed  between  two  quantities,  it  indi- 
cates that  the  first  is  to  be  multiplied  by  the  second.  Thus, 
a  X  b  indicates  that  a  is  to  be  multiplied  by  b.  If  a  stands 
for  7,  and  b  for  5,  then,  a  X  b  is  equal  to  7  X  5,  which  is 
equal  to  35. 

The  multiplication  of  quantities  is  also  indicated  by  simply 
writing  the  letters,  one  after  the  other ;  and  sometimes,  by 
placing  a  point  between  them ;  thus, 

a  x  b  signifies  the  same  thing  as  ab,  or  as  a.b. 

a  X  b  x  c  signifies  the  same  thing  as  abc,  or  as  a.b.c. 

§.  A  FACTOR  is  any  one  of  the  multipliers  of  a  product. 
Factors  are  of  two  kinds,  numeral  and  literal.  Thus,  in  the 
expression,  5abc,  there  are  four  factors :  the  numeral  factor, 
5,  and  the  three  literal  factors,  «,  b,  and  c. 

9.  The  SIGN  OF  DIVISION,  -f-,  is  read,  divided  by.  When 
written  between  two  quantities,  it  indicates  that  the  first  is 
to  be  divided  by  the  second, 

7.  How  is  the  sign  of  multiplication  read  ?     When  placed  between  two 
quantities,  what  does  it  indicate  ?     In  how  many  ways  may  multiplication 
be  indicated  ? 

8.  What  is  a  factor  ?     How  many  kinds  of  factors  are  there  ?     How 
many  factors  are  there  in  3abc  ? 

9.  How  is  the  sign  of  division  read  ?    When  written  between  two  quan- 
tities, what  does  it  indicate  ?     How  many  ways  are  there  of  indicating 
division  ? 


36  E  L  E  M  E  N  T  A  K  Y       A  L  G  K  B  U  A  . 

There  are  three  signs  used  to  denote  division.     Thns, 
a  ~-  b  denotes  that  a  is  to  be  divided  by  b. 
7  denotes  that  a  is  to  be  divided  by  b. 

a  |  b     denotes  that  a  is  to  be  divided  by  b. 

10.  The  SIGN  OF  EQUALITY,  =,  is  read,  equal  to.    When 
written  between  two  quantities,  it  indicates  that  they  are 
equal  to  each  other.     Thus,  the  expression,  a  -f  b  =  c,  in- 
dicates that  the  'sum  of  a  and  b  is  equal  to  c.     If  a  stands 
for  3,  and  b  for  5,  c  will  be  equal  to  8. 

11.  The  SIGN  OF  INEQUALITY,   >   < ,  is  read,  greater 
than,  or  less  than.     When  placed  between  two  quantities, 
it  indicates  that  they  are  unequal,  the  greater  one  being 
placed  at  the  opening  of  the  sign.     Thus,  the  expression, 
a  >  b,  indicates  that  «  is  greater  than  b ;  and  the  expres- 
sion, c  <  d,  indicates  that  c  is  less  than  d. 

12.  The  sign  .  • .  means,  therefore,  or  consequently. 

13.  A  COEFFICIENT  is  a  number  written  before  a  quan- 
tity, to  show  how  many  times  it  is  taken.     Thus, 

a  +  a-\-a  +  a  +  a  =  5a, 

in  which  5  is  the  coefficient  of  a. 

A  coefficient  may  be  denoted  either  by  a  number,  or  a 
letter.    Thus,  5x  indicates  that  x  is  taken  5  times,  and  ax 

10.  What  is  the  sign  of  equality?     When  placed  between  two  quanti- 
ties, what  does  it  indicate  ? 

11.  How  is  the  sign  of  inequality  read  ?     Which  quantity  is  placed  on 
rite  side  of  the  opening  ? 

1 2.  What  does  .  • .  indicate  ? 

13.  What  is  a  coefficient?      How  many  times  is  a  taken  in  5a.     By 
-.nnt  may  a  coefficient  be  denoted?     If  no  coefficient  is  written,  what 
coefficient  is  understood  ?     In  5ar,  how  many  times  is  ax  taken?     How 
many  times  is  x  taken  ? 


D  K  F  I  N  1  T  I  O  X       OF      T  K  R  M  8  .  37 

indicates  that  x  is  taken  a,  times.  If  no  coefficient  is  writ- 
ten^  the  coefficient  1  is  understood.  Thus,  a  is  the  same 
as  la. 

14.  AN  EXPONENT  is  a  number  written  at  the  right  and 
above  a  quantity,  to  indicate  how  many  times  it  is  taken  as 
a  factor.     Thus, 

a  x  a  is  written  a2, 

a  x  a  x  a         "  a3, 

a  X  a  x  a  x  a         "  a*, 
&c.,                    &c., 

in  which  2,  3,  and  4,  are  exponents.  The  expressions  are 
read,  a  square,  a  cube  or  a  third,  a  fourth ;  and  if  we  have 
am,  in  which  a  enters  ra  times  as  a  factor,  it  is  read,  a  to 
the  mth,  or  simply  a,  mth.  The  exponent  1  is  generally 
omitted.  Thus,  a1  is  the  same  as  a,  each  denoting  that  a 
enters  but  once  as  a  factor. 

15.  A  POWER  is  a  product  which  arises  from  the  multi- 
plication of  equal  factors.     Thus, 

a  X  «  =  a2  is  the  square,  or  second  power  of  a. 

a  X  a  X  a  =  a3  is  the  cube,  or  third  power  of  a. 

a  X  a  x  a  x  a  =  a4  is  the  fourth  power  of  a. 

a  x  a  X  .  .  .  .  =  am  is  the  mih  power  of  a. 

16.  A  ROOT  of  a  quantity  is  one  of  the  equal  factors. 
Flic  radical  sign,  -y/    ,  when  placed  over  a  quantity,  indi- 
cates that  a  root  of  that  quantity  is  to  be  extracted.     The 
root  is  indicated  by  a  number  written  over  the  radical  sign, 

14.  What  is  an  exponent?     In  a3,  how  many  times  is  a  taken  as  a  fac- 
tor?    When  no  exponent  is  written,  what  is  understood? 

1 5.  What  is  a  power  of  a  quantity  ?     What  is  the  third  power  af  2  ? 
Of  4  »     Of  6  ? 

16.  What  is  the  root  of  a  quantity?     What  indicates  a  root?     What 
indicates  the  kind  of  root?     What  is  the  index  of  the  square  root?    Of 
the  cube  root  ?     Of  the  mih  root  ? 


38  ELEMENTARY      ALGEBRA.. 

called  an  index.     When  the  index  is  2,  it  is  generally  omit- 
ted.    Thus, 

^/a,    or  y/a,  indicates  the  square  root  of  a. 
$/a    indicates  the  cube  root  of  a. 
\/a    indicates  the  fourth  root  of  a. 
'H/a    indicates  the  with  root  of  a. 

17.  An  ALGEBRAIC  EXPRESSION  is  a  quantity  written  in 
algebraic  language.     Thus, 

j  is  the  algebraic  expression  of  three  tunes 
(      the  number  denoted  by  a ; 
2  1  is  the  algebraic  expression  of  five  times 
(      the  square  of  a ; 

is  the  algebraic  expression  of  seven  times 
the  the  cube  of  a  multiplied  by  the 
(      square  of  b ; 

( is  the  algebraic  expression  of  the  differ- 
Sa  —  5b<      ence  between  three  times  a  and  five 
(      times  b ; 

is  the  algebraic  expression  of  twice  the 
.       square  of  ff,  diminished  by  three  times 
the  product  of  a  by  £>,  augmented  by 
four  times  the  square  of  b. 

18.  A  TERM  is  an  algebraic  expression  of  a  single  quan- 
tity.    Thus,  3a,  2a£,  —  5a2£2,  are  terms. 

19.  The  DEGREE  of  a  term  is  the  number  of  its  literal 
factors.     Thus, 

j  is  a  term  of  the  first  degree,  because  it  contains  but 
(      one  literal  factor. 

17.  WThat  is  an  algebraic  expression 

18.  What  is  a  term? 

19.  What  is  the  degree  of  a  term?  What  determines  the  degree  of  a  term? 


DEFINITION      OF      TEKMS.  39 

_  2  j  is  of  the  second  degree,  because  it  contains  two  lite- 
(      ral  factors. 

is  of  the  fourth  degree,  because  it  contains  four  literal 
factors.  The  degree  of  a  term  is  deteraiined  by 
the  sum  of  the  exponents  of  all  its  letters. 

20.  A  MONOMIAL  is  a  single  term,  unconnected  with  any 
other  by  the  signs  +  or  —  ;  thus,  3a2,  3  J3a,  are  monomials. 

21.  A  POLYNOMIAL  is  a  collection  of  terms  connected 
by  the  signs  +  or  —  ;  as, 

3a  -  5,  or,  2a3  —  3b  +  4bz. 

22.  A  BINOMIAL  is  a  polynomial  of  two  terms ;  as, 

a  +  b,  3a2  —  c2,  Gab  —  c2. 

23.  A  TRINOMIAL  is  a  polynomial  of  three  terms ;  as, 

abc  —  a3  4-  c3,  ab  —  gh  —  f. 

24.  HOMOGENEOUS  TERMS  are  those  which  contain  the 
same  number  of  literal  factors.     Thus,  the  terms,  abc,  —  a3, 
4-  c3,  are  homogeneous ;  as  are  the  terms,  a£>,  —  gh. 

25.  A  POLYNOMIAL  is  HOMOGENEOUS,  when  all  its  terms 
aio  homogeneous.    Thus,  the  polynomial,  abc  —  a3  +  c3,  is 
homogeneous  ;  but  the  polynomial,  ab  —  gh  —  f  is  not  ho- 
mogeneous. 

i£6.     SIMILAR  TERMS  are  those  which  contain  the  same 
literal  factors  affected  with  the  same  exponents.     Thus, 

lab  -f-  Bab  —  2ab, 

20.  What  is  a  monomial  ? 

21.  What  is  a  polynomial? 
9.2.  What  is  a  binomial  ? 

23.  What  is  a  trincmial? 

24.  What  are  homogeneous  terms  ? 

*>.5.  When  is  a  polynomial  homogeneous? 

26.  What  are  similar  terms? 


40  ELEMENTARY      ALGEBRA.. 

are  similar  terms  ;  and  so  also  are, 


but  the  terms  of  the  first  polynomial  and  of  the  last,  are  not 
similar. 

27.  THE  VrxcuLUM,  -  —  ,  the  Bar  \  ,  the  Paren- 
thesis, (  )  ,  and  the  Brackets,  [  ]  ,  are  each  used  to  con- 
nect several  quantities,  which  are  to  be  operated  upon  in  the 
same  manner.  Thus,  each  of  the  expressions, 


6-f-cxcc,         +  b 


(a  +  &  -f  c)  x 


and  [a  +  #  -f  c]  x  x, 

indicates,  that  the  sum  of  a,  £,  and  c,  is  to  be  multiplied 
by  x. 

2§.  THE  RECIPROCAL  of  a  quantity  is  1,  divided  by  that 
quantity;  thus, 

1          1          c 

a'    o~+~ft'    d' 
are  the  reciprocals  of 

A      d 

a  .    a  +  o  .    -• 

c 

29.  THE  NUMERICAL  VALUE  of  an  algebraic  expression, 
is  the  result  obtained  by  assigning  a  numerical  value  to  each 
letter,  and  then  performing  the  operations  indicated.  Thus, 
the  numerical  value  of  the  expression, 

ab  +  be  +  t7, 
•when,     a  =   1,    b  =  2,    c  =  3,    and    d  =  4,    is 

1X2  +  2X34-4  —  12; 
by  performing  the  indicated  operations. 

27.  For  what  is  the  vincular  used  ?     Point  out  the  other  ways  ir.  -whici 
this  may  be  done  ? 

28.  "What  is  the  reciprocal  of  a  quantity? 

29.  What  is  the  numerical  value  (fan  algebraical  expression? 


ALGEBRAIC      KXPKE8SION8.  41 

EXAMPLES   IX    WHITING    ALGEBRAIC   EXPRESSIONS. 

1.  Write  a  added  to  b.  Ans.  a  -f  b. 

2.  Write  b  subtracted  from  a.  Ans.  a  —  b. 

Write  the  following : 

3.  Six  times  the  square  of  a,  minus  twice  the  square  of  b. 

4.  Six  times  a  multiplied  by  £,  diminished  by  5  times  c 
cube  multiplied  by  d. 

5.  Nine  times  a,  multiplied  by  c  plus  d,  diminished  by 

8  times  b  multiplied  by  d  cube. 

6.  Five  times   a   minus   #,   plus  6  times  a  cube  into  3 
cube. 

7.  Eight  times  a  cube  into  d  fourth,  into  c  fourth,  plus 

9  times  c  cube  into  d  fifth,  minus  6  times  a  into  £,  into  c 
square.  , 

8.  Fourteen  times   a   plus   5,  multiplied  by  a  minus  b, 
plus  5  times  «,  into  c  plus  c?. 

9.  Six  times  a,  into  c  plus  </,  minus  5  times  5,  into  a  plus 
<?,  minus  4  times  a  cube  b  square. 

10.  Write  #,  multiplied  by  c  plus  <7,  plus  y  minus  </. 

11.  Write  a  divided  by  b  +  c.    Three  ways. 

12.  Write  a  —  b  divided  by  a  +  b. 

13.  Write  a  polynomial  of  three  terms;  of  four  terms;  of 
five,  of  six. 

14.  Write  a  homogeneous  binomial  of  the  first  degree;  of 
the  second ;  of  the  third ;  4th ;  5th ;  6th. 

15.  Write  a  homogeneous  trinomial  of  the  first  degree; 
.with  its  second  and  third  terms  negative;    of  the  second 

degree ;  of  the  3rd ;  of  the  4th. 

1G.  Write  in  the  same  column,  on  the  slate,  or  black-board, 
a  monomial,  a  binomial,  a  trinomial,  a  polynomial  of  four 
terms,  of  five  terms,  of  six  terms  and  of  seven  terms,  and  all 
of  the  same  degree. 


42  ELEMENT  A  11  Y       ALGEBRA 

INTERPRETATION   OF   ALGEBRAIC  LANGUAGE. 

Find  the  nunierial  values  of  the  following  expressions, 
when, 

a  =  1,    b  =  2,    c  —  3,    d  —  4. 

1.  ab  +  be.  Ans.     8. 

2.  a  •+-  be  +  d.  Ans.  11. 

3.  «c?  +  b  —  c.  Ans.     3. 

4.  aft  +  ftc  —  d.  Ans.     4. 

5.  (a  4-  ft)  c2  —  a*.  .4ns.  23. 

6.  (a  +  ft)  (a*  —  ft.)  Ans.     6. 

7.  (aft  +  ad)  c  +  d.  Ans.  22. 

8.  (aft  H    c)  (ad  —  a).  Ans.  15. 

9.  3a2ft2  -  2(a  +  d  4-  1).  ^Ins.     0. 


10.  --    x  (a  +  <?)  ^?25.  10. 

9 
«2  +    J2  +    C2  a3  _J_    J3  +    C3   _    J 

11. ' —  X •  Ans.  32. 

,    ab*  —  c  —  a3       4az  -  b  +  d3 
12.-      -—  -^  ^5.     4. 

Find  the  numerical  values  of  the  following  expressions, 
when, 

a,  =  4,    b  =  3,    c  =  2,    and  <?  =  1. 


„    a       ft 

13.  -  -   -  +  c  - 

-  (7. 

^ws.     2. 

fab       a  — 

aT\ 

14     *ii 

^I?i5.    15. 

I         o 

\   o                 o 

/ 

15.  [(a2ft  +  l)d] 

!  -   (a2ft  +  d). 

^71S.       1. 

16.  4(aftc  ) 

X   (30c3  —  ab3d3). 

Ans.  11088. 

.     a  +  ft  +  c 

abed       4a2  +  ft2  -  aT2 

Ans.  14|. 

'   a  -  ft  +  d 

aft               ftc  +  ft 

15(a+f?+ft) 

T   _l_    .             v    /Tr3/i3/>3/-73 

Ana       14  CK 

A  DDITION. 


CHAPTER  II. 

FUNDAMENTAL     OPERATIONS. 
ADDITION. 

30.  ADDITION  is  the  operation  of  finding  the  simplest 
equivalent  expression  for  the  aggregate  of  two  or  more 
algebraic  quantities.     Such  expression  is  called  their  SUM. 

When  the  terms  are  similar  and  have  like  signs. 

+ 

31.  1.  What  is  the  sum  of  a,  2a,  3a,  and  4a? 

Take  the  sum  of  the  coefficients,  and  annex  the 
literal  parts.     The  first  term,  or,  has  a  coefficient,       _|_ 
1,  understood  (Art.  13). 


2.  What  is  the  sum  of  2«#,  3a#,  6a£>,  and  ab. 
When  no  sign  is  writtten,  the  sign  -f  is  under- 
stood (Art  5). 

Add  the  following : 

(3.)  (4.)  (5.) 

a  Sab  lac  -f- 

a  lab 


-  10a 

2ab 

3ab 

Gab 

ab 

I2ab 


(6.) 


I5ab 


Sac 
I2ac 


3abc 
+  'Jabc 


30.  What  is  addition  ? 

SI.  What  is  the  rule  for  addition  when  the  terms  are  similar  and  hare 
like  sicrns  ? 


44  ELEMENTARY      ALGEBRA. 

(7.)  (8.)  (9.)  (10.) 

—  Sabc  —  Sad  —  2adf  —    Qabd 

—  2abc  —  2ad  —  Gadf  —  I5abd 

—  5abc  —  5ad  —  8adf  —  24abd 
Hence,  when  the  terms  are  similar  and  have  like  signs : 

RULE. 

Add  the  coefficients,  and  to  their  sum  prefix  the  common 
sign  ;  to  this,  annex  the  common  literal  part. 

EXAMPLES. 

(11.)                           (12.)  (13.) 

Qab  -{-    ax              8«c2  —  Sb*  lo«J3c*  —  I2abc* 

Sab  +  Sax              lac2  —  Sb2  12aft3c*  —  loabc* 

I2ab  +  4ax              Sac2  —  Qb2  ab3c*  —      abc2 

When  the  terms  are  similar  and  have  unlike  signs. 

32.  The  signs,  +  and  — ,  stand  in  direct  opposition  to 
each  other. 

If  a  merchant  writes  +  before  his  gains  and  —  before  his 
losses,  at  the  end  of  the  year  the  sum  of  the  plus  numbers 
•wall  denote  the  gains,  and  the  sum  of  the  minus  numbers 
the  losses.  If  the  gains  exceed  the  losses,  the  difference, 
which  is  called  the  algebraic  sum,  will  be  plus ;  but  if  the 
losses  exceed  the  gains,  the  algebraic  sum  will  be  minus. 

1.  A  merchant  in  trade  gained  $1500  in  the  first  quarter 
of  the  year,  §4000  in  the  second  quarter,  but  lost  $3000  in 
the  third  quarter,  and  $800  in  the  fourth  :  what  was  the  re- 
sult of  the  year's  business  ? 

1st  quarter,     +  1500  3d  quarter,     —  3000 

2cl         "  3000  4th       "  —     800 

+  4500  —  3800 

_|_  4500  —  3800   =    +  700,  or  $700  gain. 

82.  What  is  the  rule  when  the  terms  are  similar  and  hare  unlike  sitms  ? 


A  I)  I)  ITI  O  N.  45 

2.  A  merchant  in  trade  gained  $1000  in  the  first  quarter, 
and  $2000  the  second  quarter ;  in  the  third  quarter  he  lost 
$1500,  and  in  the  fourth  quarter  $1800  :  what  was  the  result 
of  the  year's  business  ? 

1st  quarter,     -f  1000  3d  quarter     —  1500 

2d             "      +  2000  4th       "            -  1800 

-f  3000  -  3300 

+  3000  —  3300   =  —  300,  or  $300  loss. 

3.  A  merchant  in  the  first  half-year  gained  a  dollars  and 
lost  b  dollars  ;  in  the  second  half-year  he  lost  a  dollars  and 
gained  b  dollars  :  what  is  the  result  of  the  year's  business  ? 

1st  half-year,  +  a  —  b 

2d        "  —  a  -f  b 

Result,  0  0 

Hence,  the  algebraic  siim  of  a  positive  and  negative  quan- 
tity is  their  arithmetical  difference,  icith  the  sign  of  the 
greater  prefixed.  Add  the  following: 

•Bab  4acbz 

Sab  —  8acbz 

—  (jab  acb2  —  2azb2c2 

5ab  —  3acbz  0 

Hence,  when  the  terms  are  similar  and  have  unlike  signs : 

I.   Write  the  similar  terms  in  the  same  column: 
B.'.  Add  the  coefficients  of  the  additive  terms,  and  also 

the  coefficients  of  the  subtractive  terms  : 
HI.  Take  the  difference  of  these  sums,  prefix  the  sign 

of  the  greater,  and  then  annex  the  literal  part. 

EXAMPLES. 

1.  What  is  the  sum  of 


46  ELEMENTARY      ALGEBRA. 

Having  written  the  similar  terms  in  the  same 
column,  we  find  the  sum  of  the  positive  coeffi- 
cients to  be  15,  and  the  sum  of  the  negative 
coefficients  to  be  —  16 :  their  difference  is  —  1 ; 
hence,  the  sum  is  —  azb3. 


2.  What  is  the  sum  of 

5a25  —  3a2S  +  4a25  —  Qa*b  —  «2J  ?       Ans. 

3.  What  is  the  sum  of 

2a3ic2—  4a3bc2+  6a3£c2—  8a3bcz+  Ua3bc*?    Ans.  I7 

4.  What  is  the  sum  of 

—  8a?b  —  Qazb  +  Uazb?  Ans.  — 

5.  What  is  the  sum  of 

labc2  —  abc2  —  labc2  +  Sabc2  +  Gabcz?  Ans.  13abcz. 

6.  What  is  the  sum  of 

Qcb3-  5cb3—  8ac2+  20c53+  9ac2  —  24c53  ?      Ans.  +  ac  , 


To  add  any  Algebraic  Quantities. 

33.     1.  What  is  the  sum  of  3a,  55,    and    —  2c? 
Write  the  quantities,  thus, 

3a  +  5b  —  2c; 
which  denotes  their  sum,  as  there  are  no  similar  terms. 

2.  Let  it  be  required  to  find  the  sum  of  the  quantities, 
2a2       4ab 
3a2  —  Sab  +    b* 

2ab  —  5b* 

5a2  —  5ab  —  4b2 

83.  What  is  the  rule  for  the  addition  of  any  algebraic  quantities? 


A  1)  i)  I  T  I  O  .V  .  47 

From  the  preceding  examples,  we  have,  for  the  addition 
of  algebraic  quantities,  the  following 

RULE. 

I.  Write  the  quantities  to  be  added,  placing  similar  terms 
in  the  same  column,  and  giving  to  each  its  proper  sign  : 

IT.  Add  up  each  column  separately  and  then  annex  the 
dissimilar  terms  with  their  proper  signs. 

EXAMPLES. 

1.  Add  together  the  polynomials, 
3a2  —  2bz  —  4ab,  5az  —  52  -j-  2ab,  and  Sab  —  3c2  —  2&3. 

The  term  3a2  being  similar  to  C  . 
5a\  we  write  8a2  for  the  result  3* 
of  the  reduction  of  these  two  <| 

terms,  at  the  same  time  slightly      ___ IT *   ~_r". 

crossing  them,  as  in  the  first  term.   I   &a~  +    ab  —  5bz  —  3c2 

Passing  then  to  the  term  —  4ab,  which  is  similar  to 
+  2ab  and  +  3ab,  the  three  reduce  to  +  ab,  which  is 
placed  after  So2,  and  the  terms  crossed  like  the  first  term. 
Passing  then  to  the  terms  involving  b2,  we  find  their  sum 
to  be  —  5b2,  after  which  we  write  —  3c2. 

The  marks  are  drawn  across  the  terms,  that  none  of  them 
may  be  overlooked  and  omitted. 

(2.)  (3.)  (4.) 

*labc  +  9ax      Sax  -}  Sb     I2a  —  6c 
—  3abc  —  3ax      Sax  —  9b    —  3a  —  Qc 


4abc  "f  6ax  I3ax  —  Qb  Qa  —  15c 


NOTE.  —  If  a  —  5,  5  =  4,  c  =  2,   x  =  1,   what  are  the 
numerical  values  of  the  several  sums  above  found  ? 


48  ELEMENTARY        ALGEBRA. 

(5.)                             (6.)  (7.) 

9a  +  /               Qax  —    Sao  3af  +    g    +  m 

—  6a  4-  <;          —  7oKC  —    9ac  <z<7  —  So/'  —  m 

—  2o  —  /                ace  +  I7ac  ab  —    ag  +  3g 

(8.)  (9.) 

7ce  +  3a5  +  3c                 8a;2  +  Oaca;  -f  13a252c2 


—  9a5  —  9c 


(10.)  (11.) 

22/i  —  3c  —  7/  +  3/7  19a/*2  +  3a' 

-    3  A  +  8c  —  2/  —  9.7  +  5a; 


(12.)  (13.) 

7a?  —  9y  4-  52  +  3  —    g  8a  +    b 

—  x  —  3y            —  8  —    g  2a  -     b  +    c 

—  x  +    y  —  3z  +  l+7#  -3a+&             4-2<? 

—  2a  4-  6y  +  82  —  1  —    g  -  6b  —  3c  +  3d 

14.  Add  together  —  b  +  3c  —  d  —  115e  4-  6/  —  5<7,  35 
»»  2c  -  3c?  -  e  +  27/,    5c  —  8(7  +  3/  -  7^,    -  75  -  6c 
•f-  J7<^+  9e  —  5/+  H<7,    —  35  —  5<?—  2e4-6/—  9^  4-  h. 

Ans.  —  85  —  109e  +  37/  -  10^  4  A. 

15.  Add  together  the  polynomials     7«25  —  3a5c  —  852o 

—  9c3  4-  cd\    Babe  —  5crb  4-  3c3  —  452c  4-  cdz,  and  4a25 

—  8c3  4-  952c  -  3d3. 

Ans.  Qa?b  4-  5abc  —  352c  —  14c3  4-  2cc72  —  3d3. 

16.  What  is  the  sum  of,    5a25c  4-  Qbx  —  4<7/,    —  3a25c 

—  6bx  4-  1  4a/,    _  «/  +  oftc  +  2a25c,    4-  6  a/  —  85a:  4-  6a25e  ? 

ylws.  10a25c  4-  bx  +  loaf. 

17.  What  is  the  sum  of    a2^2  +  3«3m  4-5,       -  6«2;z2 
-  Qa3m  —  5,    4-  95  —  Qa?m  —  5a?n2  ? 

Ans.   -  10«2>?,2  -  IZaPm  4  95. 


ADDITION.  49 

18.  What    is    tlie    sum    of    4«3ft2c  —   IGa^x  —  9ax3d, 
f  Ga3b2c  -  QaxPd  +  I7al«,    +  lGax3d  —  a*x  —  9«3ft2c? 

Ans.  a?b~c  +  ax3d. 

19.  What  is  the  sum  of    —  Iff  +  3ft  +  4<?  -  2ft    43^ 
-  3ft  +  2ft?  ^4ws.  0. 


20.  What  is  the  sum  of,    aft  4  3xy  —  m  —  n,    — 

—  3m  4-  lln  4-  e(7,    4-  3.uy  4-  4m  —  10;z-  +  fgt 

Ans.  aft  +  cd  4-  /]?• 

21.  What  is  the  sum  of   4xy  4-  n  4-  6aaj  +  9am,   —  6<ey 
4-  6/1  —  6a#  —  8am,    2xy  —  7ft.  4-  ax  —  am?     Ans.  4-  ax. 

(22.)  (23.)  (24.) 

2 (a  +  ft)  5  (a2  -  c2)  9(c3  -  a/3) 

8  (a  4  ft)  -  4(a2  -  c2)  7(c3  —  a/3) 

2(a  4  ft)  —  l(a2  —  c2)  —  10(c3  —  a/3) 

7  (a  4-  ft)  G(c3  -  a/3) 

NOTE.  The  quantity  within  the  parenthesis  must  be 
regarded  as  a  single  quantity. 

25.  Add     3a(<72  —  7t2)   -  2a(#2  -  7i2)   4-  4a(#2  —  A2) 
f-  8a(<72  —  7i2)  —  2a(^/2  —  7i2).  Ans.  lla(^72  —  7i2). 

26.  Add     3c(a2c  —  ft2)  -  9c(a2c  -  ft2)  —  7c(a2c  —  ft2) 
4-  15c(a2c  -  ft2)  4-  c(a2c  —  ft2).  Ans.  3c(a2c  —  ft2). 

34.  In  algebra,  the  term  add  does  not  always,  as  in 
arithmetic,  convey  the  idea  of  augmentation ;  nor  the  term 
sum,  the  idea  of  a  number  numerically  greater  than  any  of 
the  numbers  added.  For,  if  to  a  we  add  —  ft,  we  have, 
a  —  ft,  which  is,  arithmetically  speaking,  a  difference  be- 
tween the  number  of  units  expressed  by  a,  and  the  number 

34.  Do  the  words  add  and  sum,  in  Algebra,  convey  the  same  ideas  aa 
in  Arithmetic.  What  is  the  algebraic  sum  of  9  and  —  4  ?  Of  8  and 

—  2  ?     May  an  algebraic  sum  be  negative  ?    What  is  the  sum  of  5  and 

—  10?     How  are  such  sums  distinguished  from  arithmetical  sums? 


50  ELEMENTARY      ALGEBRA. 

of  units  expressed  by  b.  Consequently,  this  result  is  no* 
merically  less  than  a.  To  distinguish  this  sum  from  an 
arithmetical  sum,  it  is  called  the  algebraic  sum. 


SUBTRACTION. 

35.  SUBTRACTION  is  the  operation  of  finding  the  differ- 
ence between  two  algebraic  quantities. 

36.  The  quantity  to  be  subtracted  is  called  the  Subtra- 
hend ;  and  the  quantity  from  which  it  is  taken,  is  called  the 
Minuend. 

The  difference  of  two  quantities,  is  such  a  qxiantity  as 
added  to  the  subtrahend  will  give  a  sum  equal  to  the  min- 
uend. 

EXAMPLES. 

1.  From  I7a  take  6a. 

OPEEATION. 

In  this  example,  I7a  is  the  minuend,  and  6a 


the  subtrahend:  the  difference  is  lla;  because, 
lla,  added  to  6a,  gives  I7a. 


6a 


lla 


The  difference  may  be  expressed  by  writing  the  quantities 

thus: 

17a  —  6a  =  lla; 

in  which  the  sign  of  the  subtrahend  is  changed  from  -f 
to  — . 

2.  From  I5x  take  —  9x. 

The  difference,  or  remainder,  is  such  a  quantity, 
as  being  added  to  the  subtrahend,  —  Qx,  will 
givo  the  minuend,  15x,  That  quantity  is  24x, 
and  may  be  found  by  simply  changing  the  sign 


SUBTRACTION.  51 

of  the  subtrahend,  and  adding.     Whence,  we  may  write, 
I5x  —  (—  Qx)   =  24aj. 

3.  From  Wax  take  a  —  b. 

The  difference,  or  remainder,  is  such  a  quantity,  as  added 
to  a  —  J,  will  give  the  minuend,  Wax:  what  is  that  quan 
tity? 

If  you  change  the  signs  of  both 
terms  of  the  subtrahend,  and  add, 


OPKRAT10K. 


lOax 


+  a  -  b 


Rem.  lOax  —  a  +  b 
add  +  a  —  b 

IQax 


you  have,  lOax  —  a  +  b.  Is  this 
the  true  remainder  ?  Certainly. 
For,  if  you  add  the  remainder  to 
the  subtrahend,  a  —  b,  you  obtain 
the  minuend,  lOax. 

It  is  plain,  that  if  you  change  the  signs  of  all  the  terms 
of  the  subtrahend,  and  then  add  them  to  the  minuend,  and 
to  this  result  add  the  given  subtrahend,  the  last  sum  can  be 
no  other  than  the  given  minuend ;  hence,  the  first  result  is 
the  true  difference,  or  remainder  (Art.  36). 

Hence,  for  the  subtraction  of  algebraic  quantities,  we  hava 
the  following 

BT7LE. 

I.  Write  the  terms  of  the  subtrahend  under  those  of  the 
minuend,  placing  similar  terms  in  the  same  column : 

II.  Conceive  the  signs  of  all  the  terms  of  the  subtrahend 
to  be  changed  from  +  to  —,  or  from  —  to  +,  and  then 
proceed  as  in  Addition. 


EXAMPLES  OF    MONOMIALS. 

(1.)         (2.)  (3.) 

From         Sab        6ax  Qabc 

take        2ab        Zax  tabc 

Rom.        ab        Sax  Zabo 


*  2LKMENTARY      ALGEBRA. 

(4.)  (5.)  (6.) 

From  IGaWc 

taka  9a2&26' 

Rem. 


(7.)  (8.)                     (9.) 

From    .                 Sax  4abx                  2am 

take                       8c  Qac                     ax 

Rem.         3a«  —  8c  4ate  —  9ac  2am  —  ax 


10.  From  9«2i2  take  3azb2.  Ans. 

11.  From  !Ga2xy  take  —  15a2xy.  Ans. 

12.  From  12a4?/3  take  8a4y3.  Ans. 

13.  From  19c/5cc8y  take  —  I8a5xsy.  Ans. 

14.  From  3«2Z>3  take  Sa3b2.  -4«s.   3a2i3 

15.  From  7a254  take  6«4Z»2.  Ans.  7a2^4 

16.  From  Sab2  take  «255.  ^4ws.   3«5 

17.  From  a:2?/  take  y2a7.  Ans.  x^y  —  y~x. 

18.  From  Sx^y3  take  ccy.  -4?w.   3x2y3  —  xy. 

19.  From  §cizy3x  take  a^7/2.  -4?z5.    8a2y'Jx  —  xyz. 

20.  From  9a2#2  take  —  3a2i2.  Ans. 

21.  From  14a2y2  take  —  20a2y2.  J[n«. 

22.  From  --  24a455  take  16«4£5.  Ans.    — 

23.  From    -  ISx^y*  take  —  14cc2y4.  Ans.  xzy*. 

24.  From  —  47a3#2y  take  —  5a3x2v      Ans.    —  42a3#2v. 

*  V    * 

25.  From  —  94«2x2  take  3a2ic2.  Ans.    —  giazxz. 

26.  From  a  +  cc2  take  —  y3.  ^4^5.   «  -f  x2  +  y:i. 

27.  From  a3  +  i3  take  —  a3  —  53.        ^tns    2a3  +  2b3. 

28.  From  —  16a2a;3y  take  —  19a2x3y.    Ans.   +  3a2ic3y. 

29.  Froin  a2  —  x2  take  a2  +  r«2.  Ans.    —  2#3. 


SUBTRACTION.  53 


GENERAL      EXAMPLES. 

(1.)  flj  I1') 

From  6«c  —  Sab  +    c2  6ac  —  5ab  +    c2 

take    Sac  +  Sab  +  7c  *.f~     —  Sac  —  Sab  —  7c 


rfl  .          ' 

Rem.  3ac  —  Sab  +    c2  —  7c.^,gj>g)      'Sac  —  8ab  +    c2  —  7c. 

(2.)  (3.) 

From      Qax  —  a  -f  3#2  6ya;  —  3x2  +  5b 

take        9a«  —  x  +    b2  yx  —  3     -f    a 

Rem.  —  Sax  —  a  +  x  +  2&2.  5ycc  —  3<c2  -f  3  -t-  55  —  a. 


(4.)  (5.) 

From      5a3  —  4a26  +    352c  4ad  —    cc?+3a2 

take    —  2a3  +  3a"b  —    8b*c  oab  -  4cd  +  3a2  + 


Rem.      7a3  —  Ta26  4-  1.1  l~c.  —    ab 


6.  From  a  -f-  8  take  c  —  5.  ^dns.   a  —  c  +  13. 

7.  From  6a2  —  15  take  9a2  +  30.     Ana.   —  3a2  —  45. 

8.  From  Qxy  —  8«2c3  take  —  *Ixy  — 

Ans. 

9.  From  a  +  c  take  —  a  —  c.  -4?w.    2a  +  2c. 

10.  From  4(a  +  b)  take  2(o  +  b).         Ans.   2(a  +  5). 

11.  From  3  (a  +  x)  take  (a  +  a).          -4ns.   2  (a  +  a). 

12.  From  9(a2  —  a2)  take  —  2(a2  —  cc2). 

^na.    11  (a2  -  x2) 

13.  From  6a2  —  1552  take  —  3a2  +  9i2. 

Ans.    9a2  —  2452. 

14.  -From  3am  —  2bn  take  am  —  2bn.  Ans.    2am. 

15.  From  9c2m2  —  4  take  4  —  7c2?n2.   -4n*.  16c2;n2  -  8. 

16.  From  Gam  -f  y  take  3a»i  —  x.  Ans    Sam  +  x  +  y. 

17.  From  3«jc  take  3a«  —  y.  Ans.    +  y. 


54:          ELEMENTARY   ALGEBRA. 

18.  From  —  If  +  3m  —  8x  take  —  6/  —  5m  —  2x  + 
3d  +  8.  Ans.    ~  f  +  Sm  —  6x  —  3d  —  8. 

19.  From  —  a  —  5b  +  7c  -f-  d  take  4ft  —  c  -f  2d  +  2&. 

Ans.    —  a  —  9ft  +  8c  —  d  —  27c. 

20.  From  —  3a  +  ft  —  8c  +  7e  —  5/  +  3/i  -  7a  —  13</ 
take  &  +  2a  —  9c  +  8e  —  'Jx  +  7/  —  y  —  31  —  A: 

Ans.    —  5a  +  ft  +  c  —  e  —  12/  +  3/i  -  12y  +  3/. 

21.  From  2x  —  4a  —  25  -f-  5  take  8  —  55  +  a  +  6z. 

Ans.  —  4x  —  5a  +  3b  —  3. 

22.  From  3a  -{-  b  +  c  —  d  —  10  take  c  +  2a  —  c?. 

uins.   a  +  ft  —  10. 

23.  From  3a  +  ft  +  c  —  d  —  10  take  ft  —  19  +  3a. 

Ans.   c  —  d  +  9. 

24.  From  a3  +  3&2c  +  ab2  —  abc  take  ft3  +  ab*  -  ale. 

Ans.   a3  +  3b*c  —  b\ 

25    From  12as  +  6a  —  4J  +  40  take  4ft  —  3«  -f  4z  -f 
6d  —  10.  Jbis.    8a;  +  9a  —  8ft  —  Qd  +  50. 


26.  From  2x  —  3a  +  4ft  -f  6c  —  50  take  9a  +  «  -f-  6ft 
—  6c  —  40.  ^Ins.   a;  —  12a  —  2ft  +  12c  —  10. 


27.  From  6a  —  4ft  —  12c  +  12cc  take   2a;  —  Sa  +  4ft 
—  6c.  -4n*.    14a  —  8ft  —  6c  +  lOa. 


38.  In  Algebra,  the  term  difference  does  not  always,  as 
in  Arithmetic,  denote  a  number  less  than  the  minuend.  For, 
if  from  a  we  subtract  —  ft,  the  remainder  will  be  a  -f-  b  ; 
and  this  is  numerically  greater  than  a.  "We  distinguish 
between  the  two  cases  by  calling  this  result  the  algebraic 
difference. 

38.  In  Algebra,  as  in  Arithmetic,  does  the  term  difference  denote  a 
number  less  than  the  minuend  ?  How  are  the  results  in  the  two  cases, 
distinguished  from  each  other? 


SUBTRACTION.  55 

39.  When  a  polynomial  is  to  be  subtracted  from  an  al- 
gebraic quantity,  we  inclose  it  hi  a  parenthesis,  place  the 
minus  sign  before  it,  and  then  write  it  after  the  minuend. 
Thus,  the  expression, 

6a2  —  (Sab  —  2&2  +  2fo), 

indicates  that  the  polynomial,  Sab  —  262  -f-  2ic,  is  to  be 
taken,  from  6a2.  Performing  the  operations  indicated,  by 
the  rule  for  subtraction,  we  have  the  equivalent  expression : 

6a2  —  3at>  +  2b2  —  2bc. 

The  last  expression  may  be  changed  to  the  former,  by 
changing  the  signs  of  the  last  three  terms,  inclosing  them  in 
a  parenthesis,  and  prefixing  the  sign  — .  Thus, 

6«*  —  Sab  +  252  —  2bc  =  Go?  —  (Sab  -  2b2  +  25c). 

In  like  manner  any  polynomial  may  be  transformed,  as  in- 
dicated below : 

7a3  —  8a2&  —  462c  +  Qb3  =  7a3  —  (8a*b  +  We  —  653) 

=  7a3  —  9a?b  -  (4&2c  -  Qb3). 

So,3  -  W  +  c  —  d  =  8a3  —  (762  -  c  +  d) 

=  So3  -  752  -  (—  c  +  d). 

953  —  a  +  3a2  -  d  =  Qb3  -  (a  —  3a2  -f  d) 

=  96s  —  a  -  (-  3a2  +  d). 

NOTE. — The  sign  of  every  quantity  is  changed  when  it  is 
placed  within  a  parenthesis,  and  also  when  it  is  brought  out. 

40.  From  the  preceding  principles,  we  have, 

a  —  (+  b)  =  a  —  b;  and 
a  —  (-  b)  =  a  +  b. 

39.  How  is  the  subtraction  of  a  polynomial  indicated  ?    How  is  thi* 
indicated  operation  performed  ?     How  may  the  result  be  again  put  under 
the  first  form  ?     What  is  the  general  rule  in  regard  to  the  parenthesis  ? 

40.  What  is  the  s'gn  which  immediately  precedes  a  quantity  called? 
What  is  the  sign  which  precedes  the  parenthesis  called  ?    What  is  the 


66  ELEMENTARY       ALGEBRA. 

The  sign  immediately  preceding  b  is  called  the  sign  of  the 
quantity  ;  the  sign  preceding  the  parenthesis  is  called  the 
sign  of  operation  ;  and  the  sign  resulting  from  the  combin- 
ation of  the  signs,is  called  the  essential  sign. 

When  the  sign  of  operation  is  different  from  the  sign  of 
the  quantity,  the  essential  sign  will  be  —  ;  when  the  sign  of 
operation  is  the  same  as  the  sign  of  the  quantity,  the  essen- 
tial sign  will  be  -f . 


MULTIPLICATIOlSr. 

41.  1.  If  a  man  earns  a  dollars  in  1  day,  how  much  will 
he  earn  in  6  days? 

ANALYSIS. — In  6  days  he  will  earn  six  times  as  much  as  in 
1  day.  If  he  earns  a  dollars  in  1  day,  in  6  days  he  will  earn 
6  a  dollars. 

2.  If  one  hat  costs  d  dollars,  what  will  9  hats  cost  ? 

A.ns.  9d  dollars. 

3.  If  1  yard  of  cloth  costs  c  dollars,  what  will  10  yards 
cost?  •  Ans.  lOc  dollars. 

4.  If  1  cravat  costs  b  cents,  what  will  40  cost  ? 

Ans.  40&  cents. 

5.  If  1  pair  of  gloves  coats  b  cents,  what  will  a  pairs 
cost? 

ANALYSIS. — If  1  pair  of  gloves  cost  b  cents,  «  pairs  will 
cost  as  many  tunes  b  cents  as  there  are  units  in  a :  that  is, 
b  taken  a  times,  or  ab ;  which  denotes  the  product  of  b 
by  a,  or  of  a  by  b. 

resulting  sign  called?  "When  the  sign  of  opeiation  is  different  from  the 
sign  of  the  quantity,  what  is  the  essential  sign  ?  When  the  sign  of  ope- 
ration is  the  same  as  the  sign  of  the  quantity,  what  is  thr.  essential  sign  ? 
41.  What  is  Multiplication?  What  is  the  quantity  to  be  multiplied 
called?  What  is  that  called  by  which  it  is  multiplied?  What  ia  the 
result  called? 


MULTIPLICATION.  57 

MULTIPLICATION  is  the  operation  of  finding  the  product 
of  two  quantities. 

The  quantity  to  be  multiplied  is  called  the  Midtiplicand  ; 
that  by  which  it  is  multiplied  is  called  the  Multiplier ;  and 
the  result  is  called  the  Product.  The  Multiplier  and  Multi- 
plicand are  called  Factors  of  the  Product. 

6.  If  a  man's  income  is  3a  dollars  a  week,  how  much  will 
he  receive  in  45  weeks  ? 

3a  X  46  =  I2ab. 

If  we  suppose  a  =  4  dollars,  and  b  =  3  weeks,  the  pro- 
duct will  be  144  dollars. 

NOTE. — It  is  proved  in  Arithmetic  (Davies'  School,  Art.  48. 
University,  Art.  50),  that  the  product  is  not  altered  by  chang- 
ing the  arrangement  of  the  factors ;  that  is, 

I2ab  =  axbxl2   =  bxaxl2  =  axl2xb. 

MULTIPLICATION    OF    POSITIVE    MONOMIALS. 

42.     Multiply  3a2b2  by  2a2b.     "We  write, 

3a?b2  X  2«2&   =   3   X  2   X  0?  X  a?  X  b2  X  b 
—  3  x  2  a  a  a  a  b  b  b; 

in  which   a   is  a  factor  4  times,  and   b  a  factor  3  times ; 
hence  (Art.  14), 

3«262  X  2a?b  =  3  X  2a*b3  =  6a4i3, 

in  which  ice  multiply  the  coefficients  together,  and  add  the 
exponents  of  the  like  letters. 

The  product  of  any  two  positive  monomials  may  be  found 
in  like  manner;  hence  the 

RULE. 

I.    Multiply  the  coefficients  togetJier  for  a  new  coefficient: 
II.    Write  after  this  coefficient  all  the  letters  in  both  mono- 
id. V^hat  is  the  rule  for  multiplying  one  monomial  by  another  ? 
3* 


58  ELEMENTARY      ALGEBRA. 

mials,  giving  to  each  letter  an  exponent  equal  to  the  sum  of 
its  exponents  in  the  two  factors. 

EXAMPLES. 

1.  8a?bcz  X  labd2  =  56a3b2c2d*. 

2.  2la3b2cd  X  8abc3  —  168«453c4<?. 
3  4abc  X  Idf  —  2Sabcdf. 

(4.)  (5.)  (6.) 

Multiply          3a2b  I2a'2x  Gxyz 

by  2azb 


(7.)  (8.)  (9.) 

a*2-rp*i  *^/77j2/i3 

vJ\l  OCtt/    t/ 


10  Multiply  5a3b2xz  by  6c5ic6.  ^4ns. 

11.  Multiply  10a455c8  by  lacd. 

12.  Multiply  36a857c6c?5  by  20a£2c3d4. 

13.  Multiply  5am  by  3a5". 

14.  Multiply  3am53  by  6azbn. 

15.  Multiply  6amin  by  9a567.  Ans.   54am+5Z»n+". 

16.  Multiply  5ambn  by  2a^b^. 

17.  Multiply  5amb2c2  by  2a5nc. 

18.  Multiply  Ga2bmcn  by  3a3J2c2. 

19.  Multiply  20a5b5cd  by  I2a2x2y. 

20.  Multiply  14a456<?4y  by  20a3c2«2y.  ^4.  280a;b6c2d4x2y2. 

21.  Multiply  8a3i3y4  by  1a*bxy5.  Ans.   56a~b*xy9. 

22.  Multiply  I5axyz  by 


MULTIPLICAT1C   N . 


59 


23.  Multiply  64«3w5ar1?/2  by  8a5V.  A.  512a4J2c3m5sc1ya. 

24.  Multiply  9«2£W3  by  12a3J*c6.     Ans.    W8a5beced3. 

25.  Multiply  216a£W8  by  3a3Z>V.     ^4ws.    648a4#W8 

26.  Multiply  70a8£W2/«  by  12 


MULTIPLICATION    OF    POLYNOMIALS. 

43.     1.  Multiply  a  —  b  by  c. 

It  is  required  to  take  the  difference  a  —  b 

between    a  and  5,    c   times  ;  or,  to  c 

take  c,  a  —  6  times. 

As  we  can  not  subtract  b  from  c, 
we  begin  by  taking  «,  c  times,  which 
is  ac  ;  but  this  product  is  too  large 
by  b   taken    c  times,  which  is   be  ;          56  —  21   =  35 
hence,  the  true  product  is  ac  —  be. 

If  a,  5,  and  c,  denote  numbers,  as   a  =.  8, 
c  =  7,  the  operation  may  be  written  in  figures. 


ac  —  be 

8  - 

7 


=     5 
7 


=  3,  and 


Multiply    a  —  b    by    c  —  d. 

It  is  required  to  take  a  —  b  as 
many  times  as  there  are  units  in 
c  -  d. 

If  we  take  a  —  J,  c  times,  we 
have  ac  —  bc\  but  this  product  is 
too  large  by  a  —  b  taken  <?  times. 
But  a  —  b  taken  c?  times,  is  ad—db. 
Subtracting  this  product  from  the 
preceding,  by  changing  the  signs  of 
its  terms  (Art.  37),  and  we  have, 


a  —  b 
c  —  d 
ac—  be 

—  ad  +  bd 

ac—  be  —  ad  +  bd 


8-3 

7-2 


56  —  21 

-  16  +  6 
56  -  37  -f  6    =   2-3. 


(a  —  b)  X  (a  —  c)  =  ab  —  be  -  ad  +  bd. 


60  ELEMENTARY      A  L  G  E  B  K  A  . 

Hence,  we  have  the  following 

RULE     FOR     THE     SIGXS. 

I.  "When  the  factors  have  like  signs,  the  sign  of  their 
product  will  be  +  •' 

II.  When  the  factors  have  unlike  signs,  the  sign  of  their 
product  will  be  —  : 

Therefore,  we  say  in  Algebraic  language,  that  +  multi- 
plied by  +  ,  or  —  multiplied  by  —  ,  gives  -f-  ;  —  multi- 
plied by  -f-,  or  -f  multiplied  by  —  ,  gives  —  . 

Hence,  for  the  multiplication  of  polynomials,  we  have  the 
following 

RULE. 

Multiply  every  term  of  the  multiplicand  by  each  term  of 
the  multiplier,  observing  that  like  signs  give  +,  and  unlike 
signs  —  /  then  reduce  the  result  to  its  simplest  form. 

EXAMPLES   IN    WHICH    ALL   THE   TEEMS    AEE   PLUS. 

1.  Multiply        ....       3a2  +     4a6  +  b2 
by        .....       2a  +    5b 

6a3  +    8a2J+    2aS2 

The  product,  after  reducing,  +  I5a~b+  20a52  +  5b3 

becomes    ....       6a3  -f  23o26+  22a68  +  5b\ 


44.  NOTE.  —  It  Avill  be  found  convenient  to  arrange  the 
terms  of  the  polynomials  with  reference  to  some  letter  ;  that 
is,  to  write  them  down,  so  that  the  highest  power  of  that 
letter  shall  enter  the  first  term  ;  the  next  highest,  the 
second  term,  and  so  on  to  the  last  term. 

44.  How  are  the  terms  of  a  polynomial  arranged  with  reference  to  a 
particular  letter?  "What  is  this  letter  called  ?  I  'the  leading  letter  in  the 
Miultiplicand  and  multiplier  is  the  same,  which  will  be  the  leading  letter 
in  the  product  ? 


MULTIPLICATION.  61 

The  letter  with  reference  to  which  the  arrangemffit  is 
made,  is  called  the  leading  letter.  In  the  above  example  the 
leading  letter  is  a.  The  leading  letter  of  the  product  will 
always  be  the  same  as  that  of  the  factors. 

2.  Multiply    x2  +  2ax  +  a2    by    x  4-  a. 

Ans.  x3  4-  3.ax2  4-  3a2a;  4-  a3- 

3.  Multiply    x3  4-  y3    by    x  +  y. 

Ans.  x*  4-  xy*  4-  o?y  +  y4- 

4.  Multiply    3a62  +  6«2c2    by    3ab2  +  3o2c2. 

!c2  4-  18a4c*. 


5.  Multiply    a2b2  +  czd   by    a  +  b. 

Ans.  a3b2  +  aczd  +  a2b3 

6.  Multiply    3aa;2  +  9a53  +  cd5    by    6a2c2. 


7.  Multiply    Qla3x3  +  27a2a;  +  9ab    by    8a3cc?." 

x  +  >I2a4bcd. 


8.  Multiply    a3  +  3a2«  +  Saw;2  +  a^    by    a  +  x. 
Ans.  a* 


9.  Multiply    «2  +  y2    by    a;  +  y. 

.  x3  -f  ay2  4-  »2y  4-  y3. 


10.  Multiply  cc5  4-  xy6  -\-  lax   by    ax  4- 

4- 


11.  Multiply    a3  4-  3a2ft  4-  Sab2  +  b3    by    a  4-  b. 

Ans.  a4  4-  4a35  4-  6a252  '-f  4a53  +  b4. 


1  2.  Multiply    a3  4-  x*y  4-  »y2  +  y3    by    cc  4-.  y. 

4ns.  cc4  4-  2z3y  4-  2x-2y2  4-  2i«y3  4-  y4. 


13.  Multiply    x3  +  2x2  +  x  +  3    by    3«  4-  1. 

4ns.  3a*  4-  Ta3  4-  5x2  +  lOx  +  3. 


62  ELEMENTARY       ALGEBRA. 

« 

GENERAL      EXAMPLES. 

1.  Multiply   .......  2ax  —  Soft 

by      ........  3x  b. 

The  product      .......  Gax2—  Qabx 

becomes  after    ......  —  2abx 


reducing   ........     6axz—  llabx  +  3a52. 

2.  Multiply    a*  —  2b3   by    a  —  b. 

Ans.  as  —  2ab3  —  a*b  +  2b*. 

3.  Multiply    xz  —  3x  —  7    by    a;  —  2. 

Ans.  x3  —  5a;2  —  85  +  14. 

4.  Multiply    3a2  —  5ab  +  2b*    by    a2  —  7a&. 

^1«5.  3a4  —  26a36  +  37a2*2  -  Uab3. 

5.  Multiply    b2  +  b4  +  b6    by    62  —  1.      .4ns.  *8  -  J2. 

6.  Multiply  x*—  2x*y  +  4:X2yz—8xy3+  16y*  by  x  +  2y. 

Ans.  x5  +  32y5. 

7.  Multiply    4a;2  —  2y    by    2y.  -4ns.  8a2y  —  4y2. 

8.  Multiply    2x  -f  4y   by    2«  —  4y.     ^4/15.  4a2  —  16y2. 

9.  Multiply    x3  +  azy  +  xyz  +  y3   by   x  —  y. 

Ans.  x*  —  y*. 

10.  Multiply    a2  -f  %y  +  y2    by    cc2  —  ccy  +  y2. 

-4ns.  jc4  +  #2y2  4-  y4. 

11.  Multiply   2a2  —  Sax  +  4a^  by  5a2  —  6aa;  —  2a2. 

Ans.  10a4  —  27«3«  -f  34a2it-2  —  18(7^  —  8x*. 

12.  Multiply   3x2  —  2xy  +  5   by  x"2  +  2xy  —  3. 

Ans.  So4  +  4^y  —  4x*  -  ±x~if  +  IGay  —  15. 


13.  Multiply   3x3  +  2ar!y2  +  3y2  by   2X3  —  3z2y2  +  oy3. 
(  Gx6  -  Sa^y2  —  6^y4  + 


HS'      I5x3y3  -  QaPy*  +  lOa^y5  +  15y5. 


14.  Multiply   8ax  —  Gab  —  c  by   2ax  +  ab  +  c. 
Ans.  16a2ce2  —  4a2to  —  6a2J2  +  Gacx  —  7afto 


DIVISION.  (S3 


1«.  Mult.ply   So2  —  5b2  +  3c2  by  «2  — 
^4n«.  3a*  —  8aW  +  3a2c2 
16.  3a2  —  bbd  +    r/ 

—    5a2  +  45d  -  8c/. 

Pro.  red.  — 


17.  Multiply  amcc  —  a262  by   a'cc". 


18.  Multiply  «m  +  bn  by   am  —  5".          Ans.  azm  —  b2*. 

19.  Multiply   am  +  5"   by   «m  +  bn. 

Ans.  a?m  +  2ambn  +  52lt. 


DIVISION. 

45.  DIVISION  is  the  operation  of  finding  from  two  quan- 
tities a  third,  which  being  multiplied  by  the  second,  will 
produce  the  first. 

The  first  is  called  the  Dividend,  the  second  the  Divisor, 
and  the  third,  the  Quotient. 

Division  is  the  converse  of  Multiplication.  In  it,  we  have 
given  the  product  and  one  factor,  to  find  the  other.  The 
rules  for  Division  are  just  the  converse  of  those  for  Multi- 
plication. 

To  divide  one  monomial  by  another. 

46.  Divide    72a5  by   8a3.      The  division  is  indicated, 
thus : 

7205 

~8a?' 

The  quotient  must  be  such  a  monomial,  as,  being  multiplied 
by  the  divisor,  will  give  the  dividend.  Hence,  the  coefficient 

45.  What  is  division ?    What  is  the  first  quantity  called?    The  second? 
The  third?     What  is  given  in  division?     What  is  required ? 

46.  What  is  the  rule  for  the  division  of  monomials? 


64  ELEMENTART      ALGEBRA. 

of  the  quotient  must  be  9,  and  the  literal  part  a2 ;  fo; 
quantities  multiplied  by  8a3  will  give  72#s.     Hence, 


8«3 

The  coefficient  9  is  obtained  by  dividing  72  by  3;  and 
the  literal  part  is  found  by  giving  to  <r,  an  exponent  equal 
to  5  minus  3. 

Hence,  for  dividing  one  monomial  by  another,  we  have 
the  following 

RULE. 

I.  Divide  the  coefficient  of  the  dividend  by  the  Coefficient 
of  the  divisor,  for  a  new  coefficient : 

H.  After  this  coefficient  write  all  the  letters  of  the  dividend^ 
giving  to  each  an  exponent  equal  to  the  excess  of  its  <Fcpo~ 
ponent  in  the  dividend  over  that  in  the  divisor. 

SIGNS   IN   DIVISION. 

47.  Since  the  Quotient  multiplied  by  the  Divisor  must 
produce  the  Dividend :  and,  since  the  product  of  two  factors 
having  the  same  sign  will  be  +  ;  and  the  product  of  two 
factors  having  different  signs  will  be  —  ;  we  conclude : 

1.  When  the  signs  of  the  dividend  and  divisor  are  like, 
the  sign  of  the  quotient  will  be  4- . 

2.  When  the  signs  of  the  dividend  and  divisor  are  unlike, 
the  sign  of  the  quotient  will  be  — .     Again,  for  brevity,  we 
say, 

-f-  divided  by  +,  and  —  divided  by  — ,  give  -f- ; 
—  divided  by  +,  and  +  divided  by  — ,  give  — . 

+  ab  —  ab 

+  a  —  b 

—  ab  4-  ab 


47.  What  is  the  rule  fbr  the  signs,  in  division  ? 


DIVISION.  65 

EXAMPLES. 

(1.)  (2.) 

-f-  \8cisbzc  —  15<23cc2v 

_  =  +  2a2*.          tr-^2  =  +  3«^- 

(3.)  (4.) 


— ~  =    -  8a3. 
+    3abc 

5.  Divide  15ax*y3  by  —  Say.  u4w«.   — 

6.  Divide  84a£3a;  by  1252.  Ans.   labx. 

7.  Divide  —  36a465c2  by  9a352c.  Ans.    —  4aZ»3c. 

8.  Divide  —  99a454.c5  by  lla^x4.  Ans.    —  Qab2x. 

9.  Divide  lOSx6^3  by  54a^s.  ^Iws.   2xy5z~.' 

10.  Divide  64ce7y526.  by  —  le^y4^5.  -4n*.    —  4«yz. 

11.  Divide  —  Q6a~b6c5  by  12a2£c.  Ans.    —  8as5sc4. 

12.  Divide  —  38a*ft6<?4  by  2a3S5a'.  ^n«.    —  19a5(?3. 

13.  Divide  —  64a5£4c8  by  32a45c.  ^?zs.    —  2<zZ»3c7. 

14.  Divide  128a5a;6y7  by  IQaxy4.  Ans. 

15.  Dfvride  —  256aiJ9c8^7  by  lGa?bc6.  Ans.  — 

16.  Divide  200a8/n2;i2  by  —  50a7mn.  Ans.    —  4amn. 

17.  Divide  300a;3y422  by  6Qxy*z.  Ans. 

18.  Divide  27a552c2  by  —  9abc.  Ans.   — 

"  9,  Divide  64a3y628  by  32ory537.  Ans.   2a?yz. 

20.  Divide  —  88a556c8  by  Ila354c6.  Ans.    —  8a2£2c2. 

21.  Divide  77a4y324  by  —  Ila4y3z*.  Ans.    —  7. 

22.  Divide  84a4J2c2<7  by  —  42a452c2c?.  Ans.    —  2. 

23.  Divide  —  88a6J7c6  by  8a556c6.  Ans.    —  llab. 

24.  Divide  16x2  by  —  8x.  Ans.    —  2x. 

25.  Divide  —  88anb2  by  1  la"*b.  Ans.    —  8a*-mb. 


66  ELEMENTARY      ALGEBRA. 

26.  Divide  77am&"  by  —  11«2&3.  Ans.    —  *lam-ybn-*. 

27.  Divide  84a8£m  by  42an&9.  Ans.   2as-nbm-*. 

28.  Divide  —  88a^?  by  Sanbm.  Ans.    —  lla*-  "&«-"•. 

29.  Divide  QGabP  by  48#"5?.  Ans.   2al~nbP-*. 

30.  Divide  IG8xayb  by  12xnym.  Ans.    14xa~nyb~m. 

31.  Divide  256a53c2  by  16 


MONOMIAL  FRACTIONS. 

48.  It  follows  from  the  preceding  rales,  that  the  exact 
division  of  monomials  will  be  impossible  : 

1st.  When  the  coefficient  of  the  dividend  is  not  exactly 
divisible  by  that  of  the  divisor. 

2d.  When  the  exponent  of  the  same  letter  is  greater  in 
the  divisor  than  in  the  dividend. 

3d.  When  the  divisor  contains  one  or  more  letters  not 
found  in  the  dividend. 

In  either  case,  the  quotient  will  be  expressed  by  a  fraction. 
A  fraction  is  said  to  be  in  its  simplest  form,  when  the 
numerator  and  denominator  do  not  contain  a  common  factor. 
For  example,  12a452cc7,  divided  by  8a25c2,  gives 


which  may  be  reduced  by  dividing  the  numerator  and  de- 
nominator by  the  common  factors,  4,  a2,  #,  and  c,  giving 


Also, 


48.  Under  what  circumstances  will  the  division  of  monomials  be  im- 
possible ?  How  will  the  quantities  then  be  expressed  ?  How  is  a  mono- 
mial fraction  reduced  to  its  simplest  form  ? 


2c 
5a 


DIVISION.  67 

Hence,  for  the  reduction  of  a  monomial  fraction  to  its  sim- 
plest form,  we  have  the  following 


KULE. 


Suppress  every  factor,  whether  numerical  or  literal,  that 
is  common  to  both  terms  of  the  fraction  ;  the  result  will  be 
the  reduced  fraction  sought. 


EXAMPLES. 

(1.)  (20 

:  and     _  ...  .  m  = 


Sbce 
(3.)  (4.) 


2« 
also»  32  -          :   and  'T  - 


2a 
2mn 


5.  Divide  49«2&2c6  by  14a3*c4.  Ans. 

,    -r..  ., 

6.  Divide  6amn  by  3aoc.  ^3.^5. 

be 

7.  Divide  ISaWmn2  by 

8.  Divide  28a!Wd8  by 

9.  Divide  72aW  by 

10.  Divide  100a8#5a»nn  by  25a?b4d.         Ans. 

d 

11.  Divide  96a668cW  by  I5a?cxy.         Ans. 

25xy 

12.  Divide  85m2n3/a;2?/3  by  ISam^/".        ^Ins. 


13.  Divide  127d3o;2y2  by 


127 


68  ELEMENTARY      ALGEBRA. 

49.  In  dividing  monomials,  it  often  happens  that  the 
exponents  of  the  same  letter,  in  the  dividend  and  divisor, 
are  equal  ;  in  which  case  that  letter  may  not  appear  in  the 
quotient.  It  might,  however,  be  retained  by  giving  to  it  the 
exponent  0. 
If  we  have  expressions  of  the  form 

a     a2     a3     a*     a5      . 
a'    a2'.  a3'    ^'    ^'          ' 
and  apply  the  rule  for  the  exponents,  we  shall  have, 


_  a  _  a-    _  a        c> 

a  a  a3 

But  since  any  quantity  divided  by  itself  is  eqiial  to  1,  it  fol- 
lows that, 

-  =  a°  =  1,   —,  =  a2-2  =  a°  =  1,  &c.  ; 
a  a2 

or,  finally,   if  we  designate  the  exponent  by  m,  we  have, 

/"flft 

—  =  am~m  =:  a°  =  1  ;  that  is, 
am 

The  0  power  of  any  quantity  is  equal  to  1  :  therefore, 
Any  quantity  may  be  retained  in  a  term^  or  introduced 
into  a  term,  by  giving  it  the  exponent  0. 

EXAMPLES. 
1.  Divide  6«262c*  by  2a2£2. 

ftt&lfle* 

=  3c*. 


2.  Divide  8a453c5  by  —  4a453c.   ^tna.  —  2a°6°c4  =  —  2c4. 

3.  Divide  —  32m3>i2a2y2  by  ±mzri*xy. 

Ans.   —  8m°n°xy  =   —  Sxy. 

49.  When  the  exponents  of  the  same  letter  in  the  dividend  and  divisor 
are  equal,  what  takes  place  ?  May  the  letter  still  be  retained  ?  Wi  Ji 
what  exponent?  What  is  the  zero  power  of  any  quantity  equal  to? 


DIVISION.  69 

4.  Divide  —  OGa'&V1  by  —  24a4&5.  Ans.  4a°J°cn  =  4c*. 

5.  Introduce  a,  as  a  factor,  into  6b5ci.  Ans.  6a°b5c*. 

6.  Introduce  ab,  as  factors,  into  Oc5^".      Ans.  9a°b°c5dn. 

7.  Introduce  ale,  as  factors,  into  Sd\fm.     A.  8a°i°c°f?4/m. 

5O.  "When  the  exponent  of  any  letter  is  greater  in  the 
divisor  than  it  is  in  the  dividend,  the  exponent  of  that  letter 
in  the  quotient  may  be  written  with  a  negative  sign.  Thus, 

—  =   —',    also,    —  =  a"-6  =  a"3,    by  the  rule; 
a5        a3  a° 

hence,  a~3  =  -5- 

Since,         a~3  =  — =.     we  have,    b  x  a~3  =  — =; 
a3  a3' 

that  is,  «  in  the  numerator,  with  a  negative  exponent,  is 
equal  to  a  in  the  denominator,  with  an  equal  positive  ex- 
ponent; hence, 

Any  quantity  having  a  negative  exponent,  is  equal  to  the 
reciprocal  of  the  same  quantity  icith  an  equal  positive  ex- 
ponent. 

Hence,  also, 

Any  factor  may  be  transferred  from  the  denominator  to 
the  numerator  of  a  fraction,  or  the  reverse,  by  changing  the 
sign  of  its  exponent. 

EXAMPLES. 

1.  Divide  32a25c  by  16«5i2. 

16a5b2  a3b 

60.  When  the  exponent  of  any  letter  in  the  divisor  is  greater  than  in 
the  dividend,  how  may  the  exponent  of  that  letter  be  written  in  the  quo- 
tient ?  What  is  a  quantity  with  a  negative  exponent  equal  to  ?  How 
may  a  factor  be  transferred  from  the  numerator  to  the  denominator  of  a 
frn.-tion  ? 


70  ELEilENTARY      ALGEBRA. 


2- 


3.  Reduce  -rr-ns  *  Ans.  --  ,  or 

5  la;4?/3  3    '         3 

4.  In  5ay~3x~z,  get  rid  of  the  negative  exponents. 

5a 
Ans.  -— 


K    T  .. 

5.  In       _       6  ,  get  rid  of  the  negative  exponents. 

4a?i8 

Ans.     -  — 
3x2 

15a~yc~4d~s 

6.  In  -  -  —  -  —  i  ,  get  rid  of  the  negative  exponents. 

-3-5~2' 


—  — 

7.  Reduce  -^^r      ^n*.  -     -^  -,  or 

8.  Reduce  72a552  -f-  8a653.  -4wa.  Oa-1^-1.  or  — 


9.  In          _2      l    ,  get  rid  of  the  negative  exponents. 


_ 

10.  Reduce  --  --j  —  =-  •  ^Ins.  3aJ2c2. 

—       ~6~ 


To  divide  a  polynomial  by  a  monomial. 

51.     To  divide  a  polynomial  by  a  monomial  : 

Divide  each  term  of  the  dividend,  separately,  by  the 

divisor  ;  the,  algebraic  sum  of  the  quotients  will  be  the  quo- 

tient sought. 

EXAMPLES. 

1.  Divide  3a252  -  a  by  a.  Ans.  3ab2  —  1. 

61.  How  do  you  divide  a  polynomial  by  a  monomial  ? 


DIVISION.  71 


2.  Divide  5a3J2  —  25a"£2  by  5aW.  Ans.   1  —  5a. 

3.  Divide  35a262  —  25ab  by  —  Sab.  Ans.    —  lab  +  5. 

4.  Divide  10«5  —  I5ac  by  5o.  -4ra.    25  —  3e. 

5.  Divide  Gab  —  Sax  +  4a2y  by  2a. 


6.  Divide  —  15aa2  +  6a3  by  —  3x.      Ans.   5ax  —  2x2. 

7.  Divide  —  2  lay2  +  35a262y  -  7c2y  by  -  7y. 

^4?i5.   3xy  —  5a252  -f-  c2. 

8.  Divide  40a85*  +  8a457  —  32a4**c*  by  8a*b4. 

Ans.  5a*  +  53  —  4c<. 

DIVISION    OF    POLYNOMIALS. 

52.     1.  Divide  —  2a  +  6a2  —  8  by  2  +  2a. 

Dividend.     Divisor. 
6a2  —  2a  —  8  |  2a  +  2 
6a3  +  6a  3a  —  4     Quotient. 

—  8a  —  8 

—  8a  —  8 

0  Remainder. 

We  first  arrange  the  dividend  and  divisor  with  reference 
to  a  (Art.  44),  placing  the  divisor  on  the  left  of  the  dividend. 
Divide  the  first  term  of  the  dividend  by  the  first  term  of 
the  divisor  ;  the  result  will  be  the  first  term  of  the  quotient, 
which,  for  convenience,  we  place  under  the  divisor.  The 
product  of  the  divisor  by  this  term  (6«2  -f-  6«),  being  sub- 
tracted from  the  dividend,  leaves  a  new  dividend,  which  may 
be  treated  hi  the  same  way  as  the  original  one,  and  so  on  to 
the  end  of  the  operation. 

52.  What  is  the  rule  for  dividing  one  polynomial  by  another  ?  When 
is  the  division  exact  ?  When  is  it  not  exact  ? 


72  ELEMENTARY      ALGEBRA. 

Since  all  similar  cases  may  be  treated  in  the  same  way,  we 
Lave,  for  the  division  of  polynomials,  the  following 


K  U  L  E  . 

I.  Arrange  the  dividend  and  divisor  with  reference  to  the 
same  letter: 

II.  Divide  the  first  term  of  the  dividend  by  the  first  term 
of  the  divisor,  for  the  first  term  of  the  quotient.    Multiply 
the  divisor  by  this  term  of  the  quotient,  and  subtract  the 
product  from  the  dividend: 

m.  Divide  the  first  term  of  the  remainder  by  the  first 
term  of  the  divisor,  for  the  second  term  of  the  quotient. 
Multiply  the  divisor  by  this  term,  and  subtract  the  product 
from  the  first  remainder,  and  so  on: 

TV.  Continue  the  operation,  until  a  remainder  is  found 
equal  to  0,  or  one  whose  first  term  is  not  divisible  by  that 
of  the  divisor. 

NOTE.  —  1.  When  a  remainder  is  found  equal  to  0,  the 
division  is  exact. 

2.  "When  a  remainder  is  found  whose  first  term  is  not 
divisible  by  the  first  term  of  the  divisor,  the  exact  division 
is  impossible.  In  that  case,  write  the  last  remainder  after 
the  quotient  found,  placing  the  divisor  under  it,  in  the  form 
of  a  fraction. 

SECOND     EXAMPLE. 

Let  it  be  required  to  divide 

by    4ab  —  5a2  -f  3J3. 


Wo  first  arrange  the  dividend  and  divisor  with  reference 
to  a. 


DIVISION. 

Dividend. 


73 


Divisor. 


10a*- 

t—    Sa3b  — 


—  2a2+  8a6  — 
Quotient. 


-f 


25azbz—20ab3—l5b4 
25a?b2—20ab3—15b* 


+ 


(3.) 
4-  ccy2  —  2 


+  y_ 

xy 


4- 
4- 


4 


4- 


Here  the  division  is  not  exact,  and  the  quotient  is  frac- 
tional. 


(4.) 


1  4-    a 
1  -    a 


1  -  a 


2a3  4  ,  &c. 


+  2a 

4-  2a  — 


4- 
4- 


—  2a3 


4-  2a3 

In  this  example  the  operation  does  not  terminate.    It 
be  continued  to  any  extent. 


EXAMPLES. 

1.  Divide  a2  4-  2ax  +  x2  by  a  +  x. 


Ans.  a  4- 


2.  Divide  a3  —  3a2y  4-  3ay2  —  y3  by  a  —  y, 

Ans.  a2  —  2ay  4-  J/*. 


74  ELEMENTARY      ALGEBRA. 

3.  Divide  24a2£  —  I2a*cb2  —  Qab  by    —  Gab. 

Ans.  *-  4  a  -f  2a:cb  -}-•  1. 

4.  Divide    Gas4  —  96    by    3x  —  6. 

Ans.  2x?  -f  4*2  +  8z  +  16. 

5.  Divide     a5  —  5a4a  +  10a3cc2  —  -lOa2*3  +  Soa4  —  ar" 
by  a2  — 


6.  Divide  48a^  —  76aa;-  —  64a2a;  +  I05a3  by   2x  —  3a. 

Ans.  24a;2  —  2aa;  —  35«2. 

7.  Divide    y6  —  3y4a;2  +  3y2ic4  —  a;6    by    y3  —  3y2a  + 
—  a?3.  Ans.  y3  +  3y*x  +  3?/xz  -f  a^. 

8.  Divide   64«456  -  25a2i8  by   8a253  +  5ai4. 

u4;?s.  8a2i3  —  5a54. 

9.  Divide   6a3+  28a2J  +  22a52+5J3  by 

-4«*.  2a 
10.  Divide   Qax*  +  Gaa2?/6  +  42a2a2  by  aa;  + 


11.  Divide  —  loa4  +  S1azbd  —  29aV/—  2052^  + 
8c2/2  by  3a2  -  5bd  +  cf.      Ans.  -  5a2 


12.  Divide  a^  +  aj2^2  +  y*  by  a;2  —  ;ry  +  y2. 

^L?25.  a;2  +  io//  +  y2. 

13.  Divide  ar4  —  y*   by  a;  —  y. 

Ans.  x3  4-  a^y  4-  %y~  +  2/3- 

14.  Divide   3a4—  8a2ft2+  3a2c2+  554—  352c2  by  «2-  i2. 

^^5.  3a2  —  5bz  +  3c2. 

15.  Divide  6a6  -  Sa^y2—  Ga^y4^-  6x3y24-  15a;3y3—  9.r2y4 
-f  10a;2y5  +  15y5  by  3a^  +  2a;2y2  +  3y2. 

Ans.  2x3  -  3a;V  +  5y3. 

16.  Divide   —  c2+  16a2a:2—  tabc  —  Itfbx  —  6a2J2+  Gaca: 

by   8aa;  —  Qab  —  c.  Ans.  2ax  +  ab  -}-  c. 

• 

17.  Divide    Bx*  +  4a;3y  —  4a;2  —  4cc2y2  +  16a;y  —  15    by 
2a;y  +  a;2  —  3.  An-s.  3xz  —  2a*y  -f  5. 


DIVISION.  75 

18.  Divide  x5  +  3£y5   by  x  +  1y. 

Ans.  x*  —  2x3t/  +  4ce2y2  —  8xy3  +  16y*. 

19.  Divide  3a4  —  26a3b  —  14a53  +  37a252  by  2b2  —  Sab 
3a2.  Ans.  a2  —  7«£. 

20.  Divide    a4  -  54    by    a3  +  «2#  +  «52  +  b*. 

Ans.  a  —  b. 

21.  Divide    «3  —  3a2y  +  y3    by    x  -\-  y. 

3y3 
2 


22.  Divide    1  +  2a  by  1  —  a  —  a*. 

Ans.  1  +  3a  +  4a2  +  fa3  +  ,  &c. 


76  ELEMENTARY      ALGEBRA. 


CHAPTER    LIL 

tJSEFtn    FORMULAS.     FACTORING.     GREATEST  COMMON  DIVISOR. 
LEAST  COMMON   MULTIPLE. 

USEFUL   FORMULAS. 

53.  A  FORMULA  is  an  algebraic  expression  of  a  general 
rule,  or  principle. 

Formulas  serve  to  shorten  algebraic  operations,  and  are 
also  of  muchjuse  in  the  operation  of  factoring.  When  trans- 
lated into  common  language,  they  give  rise  to  practical  rules. 

The  verification  of  the  following  formulas  affords  addi- 
tional exercises  in  Multiplication  and  Division. 

(1.) 

54.  To  form  the  square  of  a  +  b,  we  have, 

(a  +  by  =  (a  +  b)  (a  +  b)  =  ct  +  2ab  +  b\ 
That  is, 

The  square  of  the  sum  of  any  two  quantities  is  equal  to 
the  square  of  the  first,  plus  twice  the  product  of  the  first  by 
the  second,  plus  the  square  of  the  second. 

1.  Find  the  square  of  2a  -f  3b.    "We  have  from  the  rule, 
(2a  +  35)2  =  4a2  +  I2ab  +  9bz. 

53.  What  is  a  formula?    What  are  the  uses  of  formulas  ? 

64.  What  is  the  square  of  the  sum  of  two  quantities  equal  to  ? 


U  8  K  F  U  L      F  t  K  M  U  L  A  6  .  77 

2.  Find  the  square  of  5ab  +  3ac. 

Ans.  25«2&2  -f  ZQtfbc  -f-  9cc2c2. 

3.  Find  the  square  of  5a2  +  8a?b. 

Ans.  25a*  +  80a4&  -f 

4.  Find  the  square  of  6ax  +  9a2ce2. 

-f  108a3a;3  + 


(2.) 
55.     To  form  the  square  of  a  difference,  a  —  b,  we  have, 

(a  -  J)2  =  (a  -  b)  (a  -  b)  =  a2  -  2ab  +  b\ 
That  is, 

The  square  of  the  difference  of  any  two  quantities  is 
equal  to  the  square  of  the  first,  minus  twice  the  product  of 
the  first  by  the  second,  plus  the  square  of  the  second. 

1.  Find  the  square  of  2a  —  b.    We  have, 

(2a  -  £)2  =  4«2  —  4«&  +  *2- 

2.  Find  the  square  of  4ac  —  be. 

Ans.  16a2c2  — 

3.  Find  the  square  of  7a2Z>2  —  12aJ3. 

Ans. 


(3.) 
56.     Multiply  a  +  b  by  a  —  b.    We  have, 

(a  +  b)  x  (a  -  5)  =  a2  —  R    Hence, 

TAe  s?m  q/"  ^wo  quantities,  multiplied  by  their  difference, 
is  equal  to  the  difference  of  their  squares. 

1.  Multiply  2c  +  b   by  2c  —  b.  Ans.  4c2  —  62. 

2.  Multiply  9ac  +  35c  by  9ac  —  35c. 

.  81a2c2  - 


55.  What  is  the  square  of  the  difference  of  two  quantities  equal  to  ? 

56.  What  is  the  sum  of  two  quantities  multiplied  by  the'.r  difference 
equal  to  ? 


78  ELEMENTARY      ALGEBRA. 

3.  Multiply  8a3  +  7aft2  by  8a3  —  7oft2. 

Ans.  64a6  —  49a2ft 

(4.) 

57.     Multiply  a2  +  ab  +  ft2  by  a  —  b.     We  have, 
(a2  +  ab  +  ft2)  (a  -  ft)  =  a3  -  ft3. 

(5.) 

5§.     Multiply  a2  --  aft  +  ft2  by  a  -f  ft.    We  have, 
(a2  -  aft  +  ft2)  (a  +  ft)  =  a3  +  ft3. 

(6.) 

59.  Multiply  together,    a  -f-  ft,    a  —  ft,    and   a2  -f  ft2. 
We  have, 

(a  +  ft)  (a  -  ft)  (a2  +  ft2)  =  a4  -  ft*. 

60.  Since  every  product  is  divisible  by  any  of  its  factors, 
each  formula  establishes  the  principle  set  opposite  its  number. 

1.  The  sum  of  the  squares  of  any  two  quantities,  plus 
twice  their  product,  is  divisible  by  their  sum. 

2.  The  sum  of  the  squares  of  any  two  quantities,  minus 
twice  their  product,  is  divisible  by  the  difference  of  the 
quantities. 

3.  The  difference  of  the  squares  of  any  two  quantities 
is  divisible  by  the  sum  of  the  quantities,  and  also  by  their 
difference. 

4.  The  difference  of  the  cubes  of  any  two  quantities  is 
divisible  by  the  difference  of  the  quantities ;  also,  by  the 
sum  of  their  squares,  plus  their  product. 

5.  The  sum  of  the  cubes  of  any  two  quantities  is  divisi- 

60.  By  what  is  any  product  divisible  ?  By  applying  this  principle,  v,-hat 
follows  from  Formula  (1)  ?  What  from  (2)?  What  from  (S)  ?  What  from 
(4)  »  What  from  (5)  ?  Wh  at  from  (6)  ? 


FACTORING.  7» 

ble  by  the  sum  of  the  quantities  ;  oho,  by  the  sum  of  their 
squares  minus  their  product. 

6.  The  difference  between  the  fourth  powers  of  any  two 
quantities  is  divisible  by  the  sum  of  the  quantities,  by  their 
difference,  by  the  sum  of  t/ieir  squares,  and  by  the  dif- 
ference of  their  squares. 


FACTOEING. 

61.  Factoring  is  the  operation  of  resolving  a  quantity 
into  factors.  The  principles  employed  are  the  converse  of 
those  of  Multiplication.  The  operations  of  factoring  are 
performed  by  inspection. 

1.  What  are  the  factors  of  the  polynomial 

ac  -f  ab  +  ad. 

We  see,  by  inspection,  that  a  is  a  common  factor  of  all 
the  terms ;  hence,  it  may  be  placed  without  a  parenthesis, 
and  the  other  parts  within ;  thus  : 

ac  +  ab  -f  ad  —  «(c  -f  b  +  d). 

2.  Find  the  factors  of  the  polynomial  azbz  +  azd  —  azf. 

Am.  az(bz  +  <?—/). 

3.  Find  the  factors  of  the  polynomial  3azb  —  6azbz  +  bzd. 

Ans.  b(3az  -  Qazb  +  bd). 

4.  Find  the  factors  of  3azb  —  9azc  —  18aVy. 

Ans.  3a2(b  —  3c  —  6xy). 

5.  Find  the  factors  of  Sa2cx  —  ISacx*  +  2acsy  ~  30a6c9. 

Ans.  2ac(4ax  —  9#2  4-  c*y  —  15a5c8). 

6.  Factor  3j&*52e  -  6a3bzd*  +  18aW. 

Ans.  6«3J2(5«c  —  d3  +  3c2). 

7.  Factor  12cW3  —  15c3J*  —  6c2f?3/. 

Ans.  3czd3(4czb  —  5cd  -  2/). 

61.  What  is  factoring  ? 


80  ELEMENTARY      ALGEBRA. 

8.  Factor  15a3bcf  —  lOabc4  —  25abcd. 

Ans.  5abc(3a-f  —  2c3  —  5d"). 

62.  When  two  terms  of  a  trinomial  are  squares,  and 
positive,  and  the  third  term  is  equal  to  twice  the  product  of 
their  square  roots,  the  trinomial  may  be  resolved  into  factors 
by  Formula  (l). 

1.  Factor    a2  -f  2ab  -f  b*.  Ans.   (a  +  b)  (a  +  b). 

2.  Factor    4a2  +  I2ab  -f  9&2. 

Ans.   (2a  +  3b)  (2a  +  3b). 

3.  Factor    9a2  +  I2ab  -f  4b2. 

Ans.   (3a  +  2b)  (3a  +  2b}. 

4.  Factor    4a;2  +  8x  +  4.         Ans.    (2x  +  2)  (2x  +  2). 

5.  Factor    9«2£2  +  I2abc  +  4c2. 

Ans.    (Sab  +  2c)  (Sab  +  2c). 

6.  Factor    lOa^y2  +  16a-y3  +  4^. 

2y2)  (4xy  -f  2y2). 


63.  When  two  terms  of  a  trinomial  are  squares,  and 
positive,  and  the  third  term  is  equal  to  minus  twice  their 
square  roots,  the  trinomial  may  be  factored  by  Formula 
(2). 

1.  Factor  az  —  2ab  +  b*.  Ans.   (a  —  b)  (a  —  b). 

2.  Factor  4a2  —  4ab  +  b*.  Ans.    (2a  —  b)  (2a  —  b). 

3.  Factor  9a2  —  6ac  +  c2.  Ans.   (3a  —  c)  (3«  —  c). 

4.  Factor  «2a;2  —  4ax  +  4.  Ans.   (ax  —  2)  (ax  —  2). 

5.  Factor  4&2  —  4.ry  +  y2.  Ans.   (2x  —  y]  (2x  —  y). 

62.  When  may  a  trinomial  be  factored  ? 

63.  When  mav  a  trinomial  be  factored  bv  this  method* 


FACTORING.  81 

6.  Factor    3Gx~  —  2-ixy  -+-  4y2. 

Ans.    (6x  —  2y)  (Qx  —  2y). 

0 

64.  Wlien  the  two  terms  of  a  binomial  are  squares  and 
have  contrary  signs,  the  binomial  may  be  factored  by 
Formula  (  3  ). 

1.  Factor    4c2  -  &.  Ans.   (2c  +  b)  (2c  —  b) 

2.  Factor    81«2c2  -  9&2c2. 

Ans.   (Qac  +  3bc)  (9«c  —  3bc). 

3.  Factor  64a4J*  —  25xzy2. 

Ans.   (8a?b2  +  5xy)  (8a?bz  —  5xy}. 

4.  Factor    25a2c2  —  9x*yz. 

Ans.    (5ac  +  3x2y)  (5ac  —  3«2y). 

5.  Factor    36a4J4c2  -  9z6. 

Ans.   (GaWc  +  3x3)  (6a262c  —  Sx3). 

6.  Factor    49a^  -  36y4.     Ans.    (7z2  +  6y2)  (7«2  —  6y2). 


65.     When  the  two  terms  of  a  binomial  are  cubes,  and 
have   contrary  signs,   the    binomial    may  be  factored  by 

Formula  (4  ). 

<» 

1.  Factor    8a3  —  c3.        Ans.  (2a  —  c)  (4«2  +  2ac  +  c2). 

2.  Factor    27«3  —  64. 

^4/zs.    (3«  —  4)  (9a2  +  12a  +  16). 

3.  Factor    a3—  64R 

^;i5.    (a  —  4i)  (a2  +  4a&  +  1652). 

4.  Factor    a3  —  .27i3.     vl»s.    (a  —  3b)  (a2  +  Sab  +  9^). 

61.  When  may  a  binomial  be  factored  ? 
65.  When  mav  a  biuomial  be  factored  by  this  method? 
4* 


82  E  L  E  M  E  N  T  A  K  T      A  L  G  K  B  B  A  . 

66.     When  the  terms  of  a  binomial  are  cubes  and  have 
like  signs,  the  binomial  may  be  factored  by  Formula  ( 5 ). 

1.  Factor    8a3  +  c3.     Ans.    (2a  +  c)  (4a2  —  2ac  +  c1). 

2.  Factor    27a3  +  64. 

Ans.    (3a  +  4)  (9a2  —  12a  +  16). 

3.  Factor    a?  +  6463. 

s.   (a  +  46)  (a2-  4a6  +  1662). 


4.  Factor    a3  +  2753.     -4«*.  (a  +  3&)  (a2  -  3aJ  +  9J2). 

67.  When  the  terms  of  a  binomial  are  4th  powers,  and 
have  contrary  signs,  the  binomial  may  be  factored  by 
Formula  (6). 

1.  What  are  the  factors  of  a4  —  54? 

Ans.  (a  +  b)  (a  —  b)  (a2  +  b2). 

2.  What  are  the  factors  of   81a*  -  16&4  ? 

Ans,    (3a  +  25)  (3a  —  26)  (9a2  -f  4J2). 

3.  What  are  the  factors  of   16a*64  —  81c4<?4? 

Ans.    (2ab  +  Zed)  (2ab  —  3cJ)  (4a262 


GREATEST    COMMON    DIVISOK, 

68.  A  COMMON  DIVISOR  of  two  quantities,  is  a  quantity 
that  will  divide  them  both  without  a  remainder.  Thus, 
3«26,  is  a  common  divisor  of  9«262c  and  3a262  —  6a?b3. 

66.  When  may  a  binomial  be  factored  by  this  method? 

67.  When  may  a  binomial  be  factored  by  this  method? 

68.  What  is  the  common  divisor  of  two  quantities  ? 


G  K  E  A  T  K  S  T      COMMON       D  I  V  I  S  O  K  .  83 

69.  A  SIMPLE  or  PRIME  FACTOR  is  one  that  cannot  be 
resolved  into  any  other  factors. 

Every  prime  factor,  common  to  two  quantities,  is  a  com- 
mon divisor  of  those  quantities.  The  continued  product  of 
any  number  of  prime  factors,  common  to  two  quantities,  is 
also  a  common  divisor  of  those  quantities. 

70.  The  GREATEST  COMMON  DIVISOR  of  two  quantities, 
is  the  continued  product  of  all  the  prime  factors  which  are 
common  to  both. 

71.  "When  both  quantities  can  be  resolved  into  prime 
factors,  by  the  method  of  factoring  already  given,  the  great- 
est common  divisor  may  be  found  by  the  following 

RULE. 

I.  Resolve  both  quantities  into  their  prime  factors  : 
II.  Find  the  continued  product  of  all  the  factors  which 
are  common  to  both  ;  it  icitt  be  the  greatest  common  divi- 
sor required. 

EXAMPLES. 

1.  Required  the  greatest  common  divisor  of  75«262c  and 
25abcL  Factoring,  we  have, 

I5a-b'zc  =  3  x  5  x  5aabbc 
25abd  =  5  x  5abd. 

The  factors,  5,  5,  a  and  b,  are  common ;  hence, 

5x5xaxb  =  25ab, 
is  the  divisor  sought. 

69.  What  is  a  simple  or  prime  factor  ?     Is  a  prime  factor,  common  to 
two  quantities,  a  common  divisor  ? 

70.  What  is  the  greatest  common  divisor  ? 

71.  If  both  quantities  can  be  resolved  into  prime  factors,  how  do  you 
find  the  greatest  common  divisor  ? 


84  ELEMENTARY      ALGEBRA. 


VERIFICATION. 

2c  ^  25ab  =  3abc 
25abd  -=-  25ab  =  cZ; 

and  since  the  quotients  have  no  common  factor,  they  cannot 
be  further  divided. 

2.  Required  the  greatest  common  divisor  of  «2  —  2ab  + 
b2  and  a2  —  b2.  ./Ins.  a  -   b. 

3.  Required  the  greatest  common  divisor  of  a2  +  2ab  -f 
62  and  a  +  b.  Ans.  a  +  b. 

4.  Required  the  greatest  common  divisor  of  «2a;2  —  4ax 
f-  4  and  ax  —  2.  Ans.  ax  —  2. 

5.  Find  the  greatest  common  divisor  of   3a?b  —  9«2c 
—  1  Sa?xy  and  &2c  —  3bc2  —  65ca^.     -4w«.  b  —  3c  —  Qxy. 

6.  Find  the  greatest  common  divisor  of  4«2c  —  4acx  and 
3«2<7  -  -  3agx.  Ans.  a(a  —  x),  or  a2  —  ax. 

1.  Find  the  greatest  common  divisor  of  4c2  —  12cx  -f-  9a;2 
and  4c2  —  9x2.  ^Iws.  2c  —  Sa1. 

8.  Find  the  greatest   common  divisor  of  x*  —  y*   and 
3.2  _  y-if  Ans.  x  —  y. 

9.  Find  the  greatest  common  divisor  of  4c2  +  4bc  -f  b2 
and  4c2  —  bz.  Ans.  2c  +  £>. 

10.  Find  the  greatest  common  divisor  of  25a2c2  — 
and  5acd2  -f  3dzx^yz.  Ans.  5ac 

NOTE.  —  To  find  the  greatest  common  divisor  of  three 
quantities.  First  find  the  greatest  common  divisor  of  two 
of  them,  and  then  the  greatest  common  divisor  between  this 
result  and  the  third. 

1.  What  is  the  greatest  common  divisor  of  4«ic2y,  IQabx2, 
and  24a«e2?  Ans.  4ax°. 

2.  Of  3ic2—  6a,  2a;3—  4#2,  and  a;2//-  2xy  ?     Ans.  x2—  2x. 


72.  When  is  one  quantity  a  multiple  of  another  ? 


LEAST      COMMON      MULTIPLE.  85 


LEAST     COilMON     MULTIPLE. 

72.  One  quantity  is  a  MULTIPLE   of  another,  when  it  can 
be  divided  by  that  other  without  a  remainder.     Thus,  8a25, 
is  a  multiple  of  8,  also  of  a2,  and  of  b.  * 

73.  A  quantity  is  a   Common  Multiple  of  two  or  more 
quantities,  when  it  can  be  divided  by  each,  separately,  with- 
out a  remainder.    Thus,  24a3x3,  is  a  common  multiple  of 
6ax  and 


74.  The  LEAST  COMMON  MULTIPLE  of  two  or  more  quan- 
tities, is  the  simplest  quantity  that  can  be  divided  by  each, 
without  a  remainder.     Thus,  12a2J2tc2,  is  the  least  common 
multiple  of  2a2cc,  4«52,  and  6«2J2tc2. 

75.  Since  the  common  multiple  is  a  dividend  of  each  of 
the  quantities,  and  since  the  division  is  exact,  the  common 
multiple  must  contain  every  prime  factor  in  all  the  quanti- 
ties ;  and  if  the  same  factor  enters  more  than  once,  it  must 
enter  an  equal  number  of  times  into  the  common  multiple. 

When  the  given  quantities  can  be  factored,  by  any  of  the 
methods  already  given,  the  least  common  multiple  may  bo 
found  by  the  following 

EULE. 

I.   Resolve  each  of  the  quantities  into  its  prime  factors  . 
II.    Take  each  factor  as  many  times  as  it  enters  any  ou6 
of  the  quantities,  and  form  the  continued  product  of  these 
factors  ;  it  will  be  the  least  common  multiple. 

73.  When  is  a  quantity  a  common  multiple  of  several  others? 

74.  What  is  the  least  common  multiple  of  two  or  more  quantities? 

75  What  does  the  common  multiple  of  two  or  more  quantities  contain, 
MS  factors?  How  may  the  least  common  multiple  be  found? 


*  The  }>i  ultiph  ol  a  quantity,  is  siraplj  \  dividend  which  will  give  an  exact  quotient 


86  ELEMENTARY      ALGEBRA. 

EXAMPLES. 

1.  Find  the  least  common  multiple  pf  12a3&2c2  and  8a253. 

I2a3b"*c2  =  2.2.3.aaabbcc. 
8a?b3  =3  2.2.2.aabbb. 

Now,  since  2  enters  3  times  as  a  factor,  it  must  enter  3 
times  in  the  common  multiple :  3  must  enter  once ;  a,  3 
times ;  £,  3  times ;  and  c,  twice ;  hence, 

2.2.2.3aaabbbcc  =  24«353c2, 
is  the  least  common  multiple. 

Find  the  least  common  multiples  of  the  following : 

2.  6a,  5«25,  and  25a£c2.  Ans.  loOa-bc2. 

3.  3a2£,  9«5c,  and  27a2»3.  Ans.  2fla2bcxz. 

4.  4a2ic2y2,  8a3tfy,  16a4y3,  and  24a5y4a;. 

5.  ax  —  bx,  ay  —  by,  and  cc2y2. 

^Lws.  (a  —  b}x.x.yy  = 

6.  a  +  b,  a2  —  52,  and  a2  +  2ab  +  J2. 

u4?w.  (a  +  6)2  (a  -  6). 

7.  3a362,  9a2a;2,  18a*y3,  3a2y\  Ans.  ISaWxty5. 

8.  8a2(a  — *),  15a5(a  -  £)2,  and  12a3(a2  —  bz). 

Ans.   120a5(a  —  J)2  (a  +  J). 


FRACTIONS.  37 


CHAPTER  IV. 


FRACTIONS. 


76.  IF  the  unit  1  be  divided  into  any  number  of  equal 
i>arts,  each  part  is  called  a  FRACTIONAL  UNIT.    Thus,  -  ,  -, 
-  ,  -  ,    are  fractional  units. 

77.  A  FRACTION  is  a  fractional  unit,  or  a  collection  of 

fractional  units.     Thus,  -  ,  -  ,  •-  ,  -=• ,  are  fractions. 
2     4     7     o 

78.  Every  fraction  is  composed  of  two  parts,  the  De- 
nominator and  Numerator.    The  Denominator  shows  into 
how  many  equal  parts  the  unit  1  is  divided ;  and  the  Nu- 
merator how  many  of  these  .parts  are  taken.     Thus,  in  the 

fraction   - ,  the  denominator  J,  shows  that  1  is  divided  into 

b  equal  parts,  and  the  numerator  <r,  shows  that  a  of  these 
parts  are  taken.  The  fractional  unit,  in  all  cases,  is  equal  to 
the  reciprocal  of  the  denominator. 

76.  If  1  be  divided  into  any  number  of  equal  parts,  what  is  each  part 
called  ? 

77.  What  is  a  fraction  ? 

78.  Of  how  many  parts  is  any  fraction  composed  ?     What  are  they 
called?     What   does  the  denominator  show?      What  the  numerator? 
What  is  the  fractional  unit  equal  to  ? 


88  ELEMENT  A  It  T      ALGEBRA. 

79.  An  EXTIKE  QUANTITY  is  one  which  contains  no 
fractional  part.  Thus,  7,  11,  a*x,  4a;2  —  3y,  are  entire 
quantities. 

An  entire  quantity  may  be  regarded  as  a  fraction  whose 

denominator  is  1.     Thus,  7  =  -,    ab  =  — • 

SO.     A  MIXED  QUANTITY  is  a  quantity  containing  both 

bx 

entire  and  fractional  parts.    Thus,    7^ ,  8^ ,  a  -\ ,    are 

c 

mixed  quantities. 

81.  Let  v    denote   any  fraction,   and   q   any  quantity 

ct 
whateyer.     From  the  preceding  definitions,  =  denotes  that 

•=-    is  taken  a  times;  also,    ~    denotes  that    T    is  taken 
0  o  o 

aq  times ;  that  is, 

aq        a 

-~  —  -   x   q',  hence, 

Multiplying  the  numerator  of  a  fraction  by  any  quan- 
tity, is  equivalent  to  multiplying  the  fraction  by  that 
quantity. 

We  see,  also,  that  any  quantity  may  be  multiplied  by  a 
fraction,  by  multiplying  it  by  the  numerator,  and  then 
dividing  the  result  by  the  denominator. 

82.  It  is  a  principle  of  Division,  that  the  same  result  will 
be  obtained  if  we    divide  the  quantity  a  by  the  product 
of  two  factors,  p  x  q,  as  would  be  obtained  by  dividing  it 

79.  What  is  an  entire  quantity  ?     When  may  it  be  regarded  as  a  frao 
tion  ? 

80.  What  is  a  mixed  quantity  ? 

81.  How  may  a  fraction  be  multiplied  by  any  quantity  ? 
82    How  may  a  fraction  be  divided  by  any  quantity  ? 


TRANSFORMATION       OF       FRACTIONS.  80 

first  by  one  of  the  factors,  p,  and  then  dividing  that  result 
by  the  other  factor,  q.     That  is, 

a         ia\  a         1 a\ 

•  —  I     )  -S-  2>    or>       -  =  I     )  •*•  PI  hence, 
pq         \pl  pq         \q/ 

Multiplying  the  denominator  of  a  fraction  by  any  quan- 
tity, is  equivalent  to  dividing  the  fraction  by  that  quantity. 

83.  Since  the  operations  of  Multiplication  and  Division 
are  the  converse  of  each  other,  it  follows,  from  the  preced- 
ing principles,  that, 

Dividing  the  numerator  of  a  fraction  by  any  quantity, 
is  equivalent  to  dividing  the  fraction  by  that  quantity  / 
and, 

Dividing  the  denominator  of  a  fraction  by  any  quantity, 
is  equivalent  to  multiplying  the  fraction  by  that  quantity. 

84.  Since  a  quantity  may  be  multiplied,  and  the  result 
divided  by  the  same  quantity,  without  altering  the  value, 
it  follows  that, 

Hoth  terms  of  a  fraction  may  be  multiplied  by  any  quan- 
tity, or  both  divided  by  any  quantity,  without  changing  the 
value  of  the  fraction. 


TRANSFORMATION     OF     FRACTIONS. 

85.  The  transformation  of  a  quantity,  is  the  operation 
of  changing  its  form,  without  altering  its  value.  The  term 
reduce  has  a  technical  signification,  and  means,  to  Trans- 
form. 

S3.  What  follows  from  the  preceding  principles  ? 

84.  What  operations  may  be  performed  without  altering  the  value  of 
a  fraction? 

85.  What  is  the  transformation  of  a  quantity  ? 


90  E  I.  £  M  E  N  T  A  R  \       ALGEBRA. 


FIRST  TRANSFORMATION. 

To  reduce  an  entire  quantity  to  a  fractional  form  having  a 
given  denominator. 

86.  Let  a  be  the  quantity,  and  b  the  given  denomi- 
nator.   We  have,  evidently,    a  =  -j- ;    hence,  the 

BULB. 

Multiply  the  quantity  by  the  given  denominator,  and 
write  the  product  over  this  given  denominator. 

SECOND   TRANSFORMATION. 

To  reduce  a  fraction  to  its  lowest  terms. 

87.  A  fraction  is  in  its  lowest  terms,  when  the  numerator 
and  denominator  contain  no  common  factors. 

It  has  been  shown,  that  both  terms  of  a  fraction  may  be 
divided  by  the  same  quantity,  without  altering  its  value. 
Hence,  if  they  have  any  common  factors,  we  may  strike 
them  out. 

EULE. 

Resolve  each  term  of  the  fraction  into  its  prime  fac- 
tors /  then  strike  out  all  that  are  common  to  both. 

The  same  result  is  attained  by  dividing  both  terms  of  the 
fraction  by  any  quantity  that  will  divide  them,  without  a 
remainder ;  or,  by  dividing  them  by  their  greatest  common 
divisor. 


86.  How  do  you  reduce  an  entire  quantity  to  a  fractional  form  having 
a  given  denominator '? 

87.  How  do  you  reduce  a  fraction  to  its  lowest  terms  ? 


TRANSFORMATION      OF      FRACTIONS.          91 


EXAMPLES. 

15<z2c2 
1.  Reduce    — — -,  to  its  lowest  terms. 


15a2c2        3.5aacc 

Factoring,  — =  — - — -; 

25acd         5.5acd 

Canceling  the  common  factors,  5,  «,  and  c,  we  have, 

15a2c*         Sac 

^ — 3  =  TT*    Ans. 
25acd         5d 

17 


2.  Reduce 

60c<W5 

3.  Reduce  ^    • 

12c5ay9 

- .    -„   ,         a5  —  ac  .        a 

4.  Reduce    — : •  Ans.   -  =  a. 

b  —  c  1 

f        T>        1  '^       *"i     +1  .  W     1 

5.  Reduce •  Ans.    -• 

n2  —  1  7i  +  l 

,,.,-.  x3  —  ax2  «? 

6.  Reduce    — : — -•  Ans. 


x2  —  2ax  +  a2  x  —  a 

9Qa3b2c  8 

7.  Reduce    — ,^  .370   •  Ans.   —-  =  —  8. 


8.  Reduce     ,„   ,-,, T---TTS-  Ans. 


a2  —  b2  A        a  +  b 

9.  Reduce    — „   .    .    T0»  jffu. 


')  r»         7  i          IO  — -*.*vv«  _ 

a2  —  2ao  +  o2  a  —  o 

5  a3  —  10a2&  +  5ab2 


10.  Reduce  ~..^. 

8a3  —  8a26  8a 

««a 

11.  Reduce 


12a4  +  6a3c2  '    4a2  + 

a2  +  2aa  +  ce2 
12.  Reduce    — ~-r—n ^ Ans. 


3(a2  -  x2)  '   3(a  - 


92  ELEMENTARY       A  L  G  E  U  R  A. . 

THIRD  TRANSFORMATION. 

To  reduce  a  fraction  to  a  mixed  quantity. 

88.  When  any  term  of  the  numerator  is  divisible  by  any 
term  of  the  denominator,  the  transformation  can  be  effected 
by  Division. 

RULE. 

Perform  the  indicated  division,  continuing  the  opei'ation 
as  far  as  possible  ;  then  write  the  remainder  over  the  deno- 
minator, and  annex  the  result  to  the  quotient  found. 

EXAMPLES. 

ax  —  a2  a? 

1.  Reduce    •  Ans.  a • 

x  x 

ax  —  ar5 

2.  Keduce •  Ans.   a  —  x. 

x 

ab  —  2a2  2a2 

3.  Reduce = •  Ans.   a =-  • 

b  o 

($,    yZ 

4.  Reduce •  An*,   a  +  x. 

a  —  x 

5.  Reduce         ~  y  •  Ans.  x2  +  xy  +  y2. 

x-  y 

10s2  -  5x  +  3  3 

6.  Reduce • —  •  Ans.   2x  —  1  +  —  • 

5x  5x 


7.  Reduce    -  -        '-T»^~    . .  4s2  -  8  + 

9x  y 

T?  ^          IBaef  —  Gbdcf  —  2ad          60      lie       2 

o»    JL«/GCIU.CG     r-      """  _  ^  -  -  •  •     -  r~  '        "         ^—  —  • 

3ac?^  d         a        3/ 

3.2  _(_  g.  ^  2 

9.  Reduce    -— —  •  Ans.  x  —  1  — 


x  +  2 


88.  How  do  you  reduce  a  fractior  to  a  mixed  quantity  " 


TRANSFORMATION      OF      FRACTIONS.          93 


10.  Reduce =-  •  Ans.  a  —  b  -\ —. 

a  +  b  a  +b 

,    _    ,         x2  +  3x  -  25  3 

11.  Reduce •  Ans.  x  -f-  Y 


-  4  'a;-4 

FOURTH    TRANSFORMATION. 

To  reduce  a  mixed  quantity  to  a  fractional  form. 

§9.  This  transformation  is  the  converse  of  the  preced- 
ing, and  may  be  effected  by  the  following 

RULE. 

Multiply  the  entire  part  by  the  denominator  of  the  frac- 
tion^ and  add  to  the  product  the  numerator  /  write  the  result 
over  the  denominator  of  the  fraction. 

EXAMPLES. 

1.  Reduce  6|  to  the  form  of  a  fraction. 
6  X  7  =  42  ;   42  +  1   =  43  ;   hence,  6}  =  ~  • 
Reduce  the  following  to  fractional  forms  : 

xz  —  (a?  —  a2)  2<e2  —  az 

v  yf  n  a 

,/JL/co. 


xx  x 

ax  +  x2  ax  —  a2 

3.  x  --  -  --  Ans. 

2a  2a 

2s-  7  ifee  -  7 

4.  5  -i  --  -  --  Ans. 

3£C  3x 

x  —  a  —  1  2a  —  x  +  1 

5.  1  ---  Ans.  --  ' 

a  a 


.    ,    „          05  —  3  lOcc2  +  4x  +  3 

6.    1  -f  2cc  --  -  --  Ans.  -  •—  -  ^— 

5£C  5x 

Sf*.  How  do  yon  reduce  a  mixed  quantity  to  a  fractional  form? 


94 

7. 
8. 
o 

ELEMENTARY 

"alb       3C  +  4 

ALGEBRA. 

16a  +  Sb  —  3c? 

4. 

Qax  +  b 
ft  _i_  S/-/7> 

8 

Qa2x  —  ab 

8 
18a2x  + 

5«J 

8  +  t%' 

4« 

f  W  WUrw  ~"|~*       Ov/Cv      V      »<^ 

^1W5. 


FIFTH    TRANSFORMATION. 

To  reduce  fractions  having  different  denominators,  to  equi- 
valent fractions  having  the  least  common  denominator. 

9O.  This  transformation  is  effected  by  finding  the  least 
common  multiple  of  the  denominators. 

13  5 

1.  Reduce  -,  -,  and  — ,  to  their  least  common  denomi- 
94  J.  +* 

nators. 

The  least  common  multiple  of  the  denominators  is  12, 
which  is  also  the  least  common  denominator  of  the  required 
fractions.  If  each  fraction  be  multiplied  by  12,  and  the  result 
divided  by  12,  the  values  of  the  fractions  will  not  be  changed. 

-    X  12  =  4,     1st  new  numerator ; 

-    X  12  =  9,     2d   new  numerator ; 
4 

—  X  12  =  5,     3rd  new  numerator ;  hence, 

i  *- 

49  o 

— ,  — ,  and  —  are  the  new  equivalent  fractions. 

90.  How  do  you  reduce  fractions  having  different  denominators,  to  equl 
valent  fractions  having  the  least  common  denominator  ?  When  the  nu- 
merators have  no  common  factor,  bow  do  you  reduce  them  ? 


TRANSFORMATION   OF   FRACTIONS.    95 


RULE. 

I.   Find  the  least  common  multiple  of  the  denominators  : 
n.   Multiply  each  fraction  by  it,  and  cancel  the  denom- 
inator : 

in.  Write  each  product  over  the  common  multiple,  and 
the  results  will  be  the  required  fractions. 

GENERAL   RULE. 

Multiply  each  numerator  by  all  the  denominators  except 
its  own,  for  the  new  numerators,  and  all  the  denominators 
together  for  a  common  denominator. 

EXAMPLES. 

ct  c 

1.  Reduce  — ^  and  =    to  their  least  common 

a2  —  b2  a  +  b 

denominator. 

The  least  common  multiple  of  the  denominators  is  (a  +  b) 
(a-b): 

l-j-  X  (a  +  b)  (a  -  b)  =  a 


az- 
c 


'—=•  X  (a  +  b)  (a  —  b)  =  c(a  —  b ;  hence, 

Ct  ~f~  0 

c(a  —  b)  ,,  .     , 

and   /„    .  v*w     '    IM   are  the  required 


(a  +  b)  (a  -  b)  (a  +  b)  (a  -  b) 

fractions. 

Reduce  the  following  to  their  least  common  denominators : 
Sx      4  12a2  45x    40     48^ 

2-  7'  6'  and  15-  An8'  Weo'-eo- 

3bz  5c3  12o     952     lOc3 

3.  a,   -,    and    -.  Ans.   _,_,_. 

,    Sx      2b  9ccc     4ab     6acd 

4.  — - ,     — ,    and    a.  Ans.    - —  ,  - —  ,  — 

2a'    3c  Qac     Qac      Gac 


96  E  L  E  M  K  X  T  ART       A  L  G  E  B  U  A  . 

3     2«  2cc  9a     Sax    12a2  +  24a? 


a;  a2  a;3 

6.  .  -  ,  TZ  --  ^,  and 


1    -a;'    (1    _a;)2»  ^    (1    _a)3 

as(l  —  a;)2    £C2(1  —  as)  x3 

Ans.  -± ~y  -77 r^,  and  7- TV 

(1  —  a;)3  '    (1  —  a;)3  (1  —  x)3 


c     c  —  b         ,       c 

>  and 


*•         M  )  )        €*.*-*  V*  — 

5a        c  a  +  0 

ac2  +  £c2        5«2c  —  5«2o  +  5abc  — 


5#2c  -f*  5abc '  5^2c  4~  5abc  '   5o&2c  +  5abc 

ex          dxz  a;3 

8. ,  — • — ,  and 

a  —  x    a  +  x  a  +  x 

:        cx(a-{-  x)    dx'2(a  —  x)         ,  x3(a  —  x) 

i-iis.       -         —  , —  ,  and  — — • 


ADDITION    OF    FBACTIOKS. 

91.  Fractions  can  only  be  added  when  they  have  a  com- 
mon unit,  that  is,  when  they  have  a  common  denominator. 
In  that  case,  the  sum  of  the  numerators  will  indicate  how 
many  times  that  unit  is  taken  in  the  entire  collection. 
Hence,  the 

KTJLE. 

L  Jteduce  the  fractions  to  be  added,  to  a  common  denom- 
inator : 

n.  Add  the  numerators  together  for  a  new  numerator^ 
and  write  the  sum  over  the  common  denominator. 

EXAMPLES. 

64  2 

1.  Add  -,  -,  and  -,  together. 
23  5 

91.  What  is  the  rule  for  adding  fractions? 


ADDITION      OF      FRACTIONS.  97 

By  reducing  to  a  common  denominator,  we  have, 

6  x  3  x  5  i=  90,  1st  numerator. 

4  x  2  x  5  =  40,  2d  numerator. 

2x3x2  =  1  2,  3d  numerator. 

2  x  3  x  5  =  30,  the  denominator. 

Hence,  the  expression  for  the  sum  of  the  fractions  becomes 

90       40       12         142  1 
30        30  +  30  Z:  ~30~; 
which,  being  reduced  to  the  simplest  form,  gives  4}J. 

CL       G  & 

2.  Find  the  sum  of  -,  -  ,  and  -• 
b    u  j 

Here,     a  X  d  x  /  =  udf  \ 

c  X  b  x  f  =  cbf  >  the  new  numerators. 

e  X  b   x  d  =  eld  ) 

and         b  x  d  x  f  =  bdf      the  common  denominator. 


<*df  ,    cbf  ,   ebd        adf  -f-  cbf  +  ebd    Al 

IIencc'  wf  +     +     =  ^—        ~  '  *  sum' 


Add  the  following  : 

3x2  2ax  2abx 

3.  a  --  r-  ,  and  b  -\  --  •  Ans.  a  +  b  -\ 


,  . 

o                       c  oc 

.     85       X              _     X  .                        X 

4.  -,    -,    and  -.  Ans.  *  +  - 

,    x  —  2         .    4«  19a;  —  14 

5.  ___   and    y.  Ans.      —  — 

x—2        ,          ,   2x  —  3  ,   lOa;  —  17 

6.  x  H  --  -  —  and    3x  -\  --  -  --    Ans:  4cc  -\  --  —  - 

3                                   4  12 

5x2          T    x  +  a  Sx3  +  ax  -+-  a* 

7.  4cc,    -—  ,    and    —  -! 


.    2«      7a;          _    2a  +  1  49o;  +  12 

8.  T,    T,    and   —g—  •  Ans.  2x  +  —^  -- 

YO*                                     //•  44:35 

9.  4«,    --,    and    2  H-  -  ylw*.   2  +  4*  +  — 

5  45 


98  ELEMENTARY       ALGEBRA. 

10.  3x  -f  ~   and   x  -  ^.  Ans.   3x  +  — 

59  45 

6b  c 

11.  ac  —  --   and    1 ,• 

8a  d 


wi/u.  -f-  Sac 
Ans.   1  -f  ac 

Sad 

±3?  -  5x  +  4 


"  4(1  +  a)'  4(1  -a)'       "2(f~^)  "  l> 


SUBTRACTION    OF    FEACTIOIfS. 

92.  Fractions  can  only  be  subtracted  when  they  have 
the  same  unit;  that  is,  a  common  denominator.  In  that 
case,  the  numerator  of  the  minuend,  minus  that  of  the  sub- 
trahend, will  indicate  the  number  of  times  that  the  common 
unit  is  to  be  taken  in  the  difference.  Hence,  the 

RULE. 

I.  Reduce  the  two  fractions  to  a  common  denomi- 
inator  : 

n.  Then  subtract  the  numerator  of  the  subtrahend  from 
that  of  the  minuend  for  a  new  numerator,  and  write  the 
remainder  over  the  common  denominator. 

EXAMPLES. 

3  2 

1.  What  is  the  difference  between  -  and  -  • 

7  8 

3        2         24        14         10          5 

_     _       __      ,_    ._      ^__     —      —      ._     _     . 

7        8   ""   56        56   "  56   ~"   28 
92.  What  is  the  rule  for  subtracting  fractions  ? 


MULTIPLICATION      OF      FRACTIONS.  9t» 


3*  —  Cl  2l7   •—  4»J* 

2.  Find  the  difference  of  the  fractions  —  r  -  and  —        —  • 

20     .  3c 

j      (cc  —  a)  x  3c  =  Sex  —  Sac  [  . 
Here,    i  .   v        .j:       rt.  0.     >  the  numerators, 

(  (2a  —  4aj)  x  20  =  4ao  —  8&c  ) 

and,  2b  X  3c  =  Qbc  the  common  denominator. 

3cx—  Sac      4ab—8bx       3cx—3ac—4:ab+8bx 
Hence,  --  -  ---  -=  -  =  --  —  •  Ans. 

Qbc  Qbo  6bc 


,    -D       •    j  ^    *•#  f  *  3x          A        39a; 

3.  Required  the  difference  of  -—  and  —  •       Ans.  —  —  • 

7  5  3o 


4.  Required  the  difference  of  5y  and  —  •         Ans.  —  - 

8        '^.  8 

5.  Required  the  difference  of  —  and  —  •         Ans.  -  —  • 

t  \s  Do 

e.  From  Z±V  subtract  ^  -  ^n*. 

x  —  y  x  +  y 


__        _ 

7.  From  -  subtract  —  -  -•         Ans.  -  —  -  --  ;-  • 

y  —  2  y2  —  s2  y2  —  22 

Find  the  differences  of  the  following  : 

3*  +  a       ,  2a;  -f  7  24«  +  8«  —  105*  —  35o 

8.  —  -r  —  and  —  -  —  •     Ans.  --  —;  --  • 

ob  8  406 

x      ,  x  —  a  ,    ex  +  bx  —  ab 

9.  3x  +  -  and  x  ---  Ans.  2x  -\  --  =  --  • 

be  be 

a  —  x         ,      a  +  a  4a5 

10.    a  -i  —  -.  -  r  and  —.  —  !  -  :•      Ans.  a 


(»       ciu.v&  f  fc  ^.m./tw#      \Af  ,.  ,j 

a  +  x)          a(a  —  05)  a*  —  or 

MULTIPLICATION    OF    FRACTIONS. 

f*  y» 

93.     Let  v  and  -^,  represent  any  two  fractions.     It  has 

O  W 

been  shown  (Art.  81),  that  any  quantity  may  be  multiplied 

93.  What  13  the  rule  for  the  multiplication  of  fractions  ? 

-* 


lUU  E  L  K  M  K  X  T  A  K  Y       A  L  G  K  B  K  A  . 

by  a  fraction,  by  first  multiplying  by  the  numerator,  and 
then  dividing  the  result  by  the  denominator. 

CL  G  CIC 

To  multiply  j-  by  -^,  we  first  multiply  by  e,  giving  —  ; 

0  Ctr  0 

then,  we  divide  this  result  by  d,  which  is  done  by  multiply- 


ac 


ing  the  denominator  by  d;  this  gives  for  the  product,  •=— ; 

OCl/ 

that  is, 


a       c         as      , 

T  X  -,  =  r^;    hence, 

b       d        bd 


RULE. 

I.  If  there  are  mixed  quantities,  reduce  them  to  a  frac- 
tional form  ;  then, 

IL  Multiply  the  numerators  together  for  a  new  numera- 
tor, and  the  denominators  for  a  new  denominator. 

EXAMPLES. 

bx  ,      c      _,.  to        a2  +  to 

1.  Multiply  a  -\ by  -•     First,    a  H =  , 

a      J  d  a  a 

,  «2  +  to        c        a2c  +  bcx 

hence,  — x  -,  =  -= —  •    Ans. 

a  a  ad 

Find  the  products  of  the  following  quantities : 

,    2x    Sab         ,  Sac 

2.  — ,  — ,  and  — j-  Ans.  9ax. 

a       c  2b 

to          a  ab  -f  to 

3.  b  -\ and  -  •  Ans.  -          —  • 

ax  x 

4.  — and  — Ans.  ^ — -— r-=- 

be  b  +  c  b2c  +  bo3 

x  +  1         ,  x  —  1  ax*  —  ax  +  xz  —1 

5.  x  -\ ,  and  7  •      Ant. • 

a  a  +  b  a2  +  ab 

ax          ,  a2  —  xz  a3  +  a2x 

6.  a  -i and •  Ans. • 

a  —  x  x  +  ar  x  +  x2 


MULTIPLICATION      OF      FRACTIONS.  101 

2d  a?  J2 

7.  Multiply         — i    by         -^ —  • 
1  J     a  —  b  3 


la  a*  —  b2        2a(ai  —  i2)         2a(a  +  b)  (a  -  b~) 

T      X      " 


a  —  J  3  3  (a  —  6)  3  (a  — 


After  indicating  the  operation,  we  factored  both  numera- 
tor and  denominator,  and  then  canceled  the  common  factors, 
before  performing  the  multiplication.  This  should  be  done, 
whenever  there  are  common  factors. 

9.  0-2  —    4/2  9,(<r.    4-    1/\ 


o.                             uy                    —  . 

x  -  y                    a 

a 

x2  —  4                   4cc 

.        4a(a;  —  2) 

S                ^       3*  4-  2 

O                                   «C    ~y^    *- 

3 

in                           Tw 

-4ws.   2a;(a  +  5). 

2a;                     (a  +  6) 

(x  -  I)2      ,          (a;  +  l)y2 

x                              nv                            /4/     . 

3.2  i 

y        by       «  -  1 

y 

r,    («2-*2)    b      i+«, 

A        a  —  x 

1  _  &        ^     a  +  x 

1  —  x 

O'v»i/                                               Oo*)/ 

£ivul.i                   -.                                                  *-t<    '/ 

Ans.  a2. 

x  —  y                       x  +  y 

2a  —  b               Ga  —  2b 

5  —  3(35 

'                                            "j              19            r»      r 

4a                   o2—  2a6 

2a& 

V2               x       y 
16.    a;  —  —     by      -  +  -• 

a4  —  v4 
Ans.  —  -Z-Z- 

102  ELEMENTARY      ALOEBKA. 


DIVISION      OF      FRACTIONS. 

P  1 

94.     Since    •-  =  p  x   -  ,    it  follows  that,  dividing  by  a 

quantity  is  equivalent  to  multiplying  by  its  reciprocal.     But 

c          d 
the  reciprocal   of  a  fraction,    -,  is   --    (Art.  28);    conse- 

Cv  G 

quently,  to  divide  any  quantity  by  a  fraction,  we  invert  the 
terms  of  the  divisor,  and  multiply  by  the  resulting  fraction. 
Hence, 

a         c        a        d        ad 

_      ,_•  T      _       T   ^       y       ^L_      . 

b        d  ~~   b        c   ~ '   be 

Whence,  the  following  rule  for  dividing  one  fraction  by 
another : 

RULE. 

I.   Reduce  mixed  quantities  to  fractional  forms  : 
II.   Invert  the  terms  of  the  divisor,  and  multiply  the 
dividend  by  the  resulting  fraction. 

NOTE. — The  same  remarks  as  were  made  on  factoring 
and  reducing,  under  the  head  of  Multiplication,  are  appli 
cable  in  Division. 

EXAMPLES. 

1.  Divide    a  —  —    by    £• 
2c  g 

b         2ac  —  b 

a =  

2c  2c 

„  b        f        2ac  —  b       fj       2acff  —  bg       . 

Hence,    a j-^-  =  -  -  x  5  =  -     --?- -.   Ans. 

2c       g  2c  f  2cf 


94.  What  is  the  rule  for  the  division  of  fractions? 


DIVISION       OF      FRACTIONS.  103 

.    ~.  . ,      2(»  +  y)    .        «-  —  y2 
2.  Divide      v         y'    by ^-- 


a  a 


+  y)  Y «_    _  2(3  +  y) 

A         o  o     — —  s* 


a  (x  +  y)  (x  -  y) 

2 


Ans, 


/>• QI 

x       y 

7«c  12  9  Is* 

3.  Let    —    be  divided  by    :—  •  Ans.  -  —  • 

5  60 


4.  Let    — —    be  divided  by    5x.  , 

7  3o 

T        a:  +  1  ,-..,.,       2a;  a;  +  1 

5.  Let    be  divided  by    —  •  Ans.   — - — 

6                                    3  4x 

x                                  x  A  2 

6.  Let    be  divided  by   -  •  Ans.   - 

x  —  I                              2  x  —  1 


c/j*  26R 

7.  Let    —    be  divided  by   ~  •  

3  30  2« 

x  —  b    .          .,    ,  ,       3ca;  x  —  I 

8.  Let    =-    be  divided  by  — •=  •  Ans. 


Divide  the  following  fractions: 

-     **-**    '       a2-4  Ans.      4X 


3                         3  *  +  2 

10 bv •  Ans.  x  +  — 

iv/'       «.9.  nX^.     I      XS         «          » A  /»• 


_            _     _  X 

4xz  (a  +  o)2 

11.    2a*(«  +  *)    by         -—=•  Ans.               '  > 

a  +  o  *x 

'--   \-  1)7/2  (x  —  I)2 

i2L  .  ^1«S. ..— — 

-  i  y3 

^2 fla.           3^  a;")  4a(a2  —  cc2) 

'    6c  +  bx    bj    4(^+^) '  4W*'    3i(c2  -  a2) 


104  ELEMENTARY      A  L  G  E  B  K  A . 


14. 

15. 

a 

—  x 

by 

1    + 

X 
X 

t 

y 

Ans. 
Ans.   x  - 

a* 

-as" 

I  —  x 
x2    by 

a  + 
2xy 

1 

i_ 

—  a;2 

• 

*  + 

1   x 

-y 

16. 

b 

—  3a 

by 

Qa  - 

-2b 

Ans. 

2a 

-  b 

• 

2ab 

62  - 

2ab 

4a 

T7 

rt 

-y4 

l-nr 

_L 

y 

Ana 

X* 

-  y2 

x'y  y       x 


18.    m2  +  1  -r  -^    by  m  H hi. 

m*      J  m 


Ans.   m  -\ 1. 

m 


IP 


(y  —  x  \  I  i 

aj  +  f-r )    by    (l  -  xf 
I  +  xyJ      J     \  1 


(x  +  2y       x\    .       /x  +  2y  a;      \ 

20.    I  —  7—^  +  -      by    I  -     —  ^  --  •  —  ) 

V*  +  y      y/         Vy         *  +  y/ 


EQUATIONS   OF  THE  FIRST  DEGREE.    105 


CHAPTER  V. 

EQUATIONS     OP     THE     FIRST     DEGREE. 

95.  AN  EQUATION  is  the  expression  of  equality  between 
two  quantities.  Thus, 

x  =  b  +  c, 

is  an  equation,  expressing  the  fact  that  the  quantity  ic,   is 
equal  to  the  sum  of  the  quantities  b  and  c. 

9O.  Every  equation  is  composed  of  two  parts,  connected 
by  the  sign  of  equality.  These  parts  are  called  members : 
the  part  on  the  left  of  the  sign  of  equality,  is  called  the  first 
member  ;  that  on  the  right,  the  second  member.  Thus,  in 
the  equation, 

x  -\-  a  =  b  —  c, 

x  +  a  is  the  first  member,  and  b  —  c,  the  second  member. 

97.  An  equation  of  the  first  degree  is  one  which  involves 
only  the  first  power  of  the  unknown  quantity ;  thus, 

Qx  +  3x  —  5   =   13;    (1  ) 
and          ax  +  bx  +  c  =.    d ;    (2) 

are  equations  of  the  first  degree. 

95.  What  is  an  equation  ? 

96.  Of  how  many  parts  is  every  equation  composed?    How  are  the 
parts  connected  ?     What  are  the  parts  called  ?     What  is  the  part  on  the 
left  called?     The  part  on  the  right  ? 

97.  What  is  an  equation  of  the  first  degree  ? 

5* 


103  ELEMENTARY       ALGEBRA. 

98.  A  NUMERICAL  EQUATION  is  one  in  which  the  ^effi- 
cients of  the  unknown  quantity  are  denoted  by  numbers. 

99.  A  LITERAL  EQUATION  is  one  in  which  the  coefficients 
of  the  unknown  quantity  are  denoted  by  letters. 

Equation  ( 1 )  is  a  numerical  equation  ;  Equation  ( 2  )  is  a 
literal  equation. 

EQUATIONS     OF    THE     FIRST     DEGREE    CONTAINING     BUT    ONE 
UNKNTOWN    QUANTITY. 

100.  The  TRANSFORMATION  of  an  equation,  is  the  opera- 
tion of  changing  its  form  without  destroying  the  equality 
of  its  members. 

101.  An  AXIOM  is  a  self-evident  proposition. 

102.  The  transformation  of  equations  depends  upon  tin 
following  axioms: 

1.  If  equal  quantities  be  added  to  both  members  of  an 
equation,  the  equality  icill  not  be  destroyed. 

2.  If  equal  quantities  be  subtracted  from  both  members 
of  an  equation,  the  equality  will  not  be  destroyed. 

3.  If  both  members  of  an  equation  be  multiplied  by  the 
same  quantity,  the  equality  will  not  be  destroyed. 

4.  If  both  members  of  an  equation  be  divided  by  the  same 
quantity,  the  equality  will  not  be  destroyed. 

5.  Like  powers  of  the  two  members  of  an  equation  are 
equal. 

6.  Like  roots  of  the  two  members  of  an  equation  are 
equal. 

98.  What  is  a  numerical  equation  ? 

99.  What  is  a  literal  equation  ? 

100.  What  is  the  transformation  of  an  equation  ? 

101.  What  is  an  axiom  ? 

102.  Name  the  axioms  on  which  the  transformation  of  an  equation 
depends. 


CLEARING     OF      FRACTIONS.  107 

103.  T\vo  principal  transformations  are  employed  in  the 
solution  of  equations  of  the  first  degree:   Clearing  of  frac- 
tions^ and  Transposing, 

CLEARING    OF    FRACTIONS. 

1.  Take  the  equation, 

2#          3«5          £C 

3~  ' '  7  "f  6   = 

The  least  common  multiple  of  the  denominators  is  12.  If 
we  multiply  both  members  of  the  equation  by  12,  each  term 
will  reduce  to  an  entire  form,  giving, 

Sx  —  Qx  +  2»  =   132. 

Any  equation  may  be  reduced  to  entire  terms  in  the  same 
manner. 

104.  Hence  for  clearing  of  fractions,  we  have  the  fol- 
lowing 

RULE. 

I.   Find  the  least  common  multiple  of  the  denominators: 
II.   Multiply  both  members  of  the  equation  by  it,  reduc- 
ing the  fractional  to  entire  terms. 

NOTE. — 1.  The  reduction  will  be  effected,  if  we  divide  the 
least  common  multiple  by  each  of  the  denominators,  and 
then  multiply  the  corresponding  numerator,  dropping  the 
denominator. 

2.  The  transformation  may  be  effected  by  multiplying 
each  numerator  into  the  product  of  all  the  denominators 
except  its  own,  omitting  denominators. 

103.  How  many  transformations  are  employed  in  the  solution  of  equa- 
tions of  the  first  degree  ?     What  are  they  ? 

104.  Give  the  rule  for  clearing  an  equation  of  fractions?     In  what  throe 
«vays  may  the  reduction  be  effected? 


108  ELEMENTAKY      ALGEBRA. 

3.  The  transformation  may  also  be  effected,  by  multiplying 
both  members  of  the  equation  by  any  multiple  of  the  de- 
nominators. 

EXAMPLES. 
Clear  the  following  equations  of  fractions  : 

1.  f  +  f  —  4  =  3.        Ans.  ?x  +  5x  —  140  =   105. 
5         7 

2.  ';  +  -  —  —  —  8.        Ans.  925  +  Qx  —  2x  =  432. 

O  J  —  / 

x      x      x       x 

3.  -  H  -----  1  --  =  20. 
•2^3        9  ^  12 

Ans.  I8x  +  12a;  —  4x  +  3*  =  720. 

•C         £C         SB 

4.  -  +  -  —  -  =  4.     .4ws.  14a;  +  lOa;  —  3ox  =  280. 
o         i         Z 


5.        —      +       =  15.     ,4w*.  15aj  —  12a;  +  lOaj  =  900. 
x  —  4       x  —  2         5 


6          " 

.  —  2a;  +  8  —  cc  -f-  2  =  10. 

x  3 

7.    -  --  h  4  =  -  •    -4ns.  5z  +  60  —  20a  =  9  —  3x. 

x       «,a5,«_19 
8'    4  ~  6  +  8  +  9  -     12' 

-4ns.  18«  —  12a;  +  9x  +  8«  =  864. 

ft  ft 

9.    -  —  -j  -\-  f  =  g.        Ans.  ad  —  be  -f  bdf  =  bdg 
o       u 


axe,       2c2ai   ,  4bo-x       5a3       <*<? 

10.    -r  --  =r-  +  4a  =  —  -  ---  75-  H  ---  30. 
b         ab  a3          b2         a 

The  least  common  multiplt  of  the  denominators  is  cr-3^2  \ 


TRANSPOSING.  109 


TRANSPOSING. 

105.  TRANSPOSITION  is  the  operation  of  changing  a  term 
from  one  member  to  the  other,  without  destroying  the 
equality  of  the  members. 

1.  Take,  for  example,  the  equation, 

5x  —  6 ;  =  8  +  2x. 

If,  in  the  first  place,  we  subtract  2x  from  both  members 
the  equality  will  not  be  destroyed,  and  we  have, 

5x  —  6  —  2x  =  8. 

Whence  we  see,  that  the  term  2ce,  which  was  additive  in 
the  second  member,  becomes  subtractive  by  passing  into 
the  first. 

In  the  second  place,  if  we  add  6  to  both  members  of 
the  last  equation,  the  equality  will  still  exist,  and  we  have, 

5x  —  6  —  2x  +  6   =   8  +  6, 

or,  since  —  6  and  +  6  cancel  each  other,  we  have, 
5x  —  2x  =  8  +  6. 

Hence,  the  term  which  was  subtractive  in  the  first  member, 
passes  into  the  second  member  with  the  sign  of  addition. 

106.  Therefore,  for  the  transposition  of  the  terms,  we 
have  the  following  » 

BULK. 

Any  term  may  be  transposed  from  one  member  of  an 
equation  to  the  other,  if  the  sign  be  changed. 

105.  What  is  transposition? 

106.  What  is  the  rule  for  the  transposition  of  the  terms  of  an  equation  ? 


110          ELEMENTARY   ALGEBRA. 
EXAMPLES. 

Transpose  the  unknown  terms  to  the  first  member,  and 
thp  known  terms  to  the  second,  in  the  following : 

1.  3x  -f  6  —  5  =  2x  —  7.   Ans.  3x  —  2x  =  —  7  —  6  -f  5. 

2.  ax  +  b  =  d  —  ex.  Ans.   ax  +  ex  =  d  —  b. 

3.  4x  —  3   =  2x  +  5.  Ans.    4x  —  2x  —   5  +  3. 

4.  Qx  +  c  —  ex  —  d.         Ans.   dx  —  ex  =   —  d  —  c. 

5.  ax  +  /  =  dx  +  b.  Ans.  ax  —  dx  —  b  —  f. 

6.  6#  —  c  =  —  02  +  5.          ^l?zs.  62;  -f  ax  =.  b  +  c. 


SOLUTION      OF      EQUATIONS. 

1O7.  The  SOLUTION  of  an  equation  is  the  operation  of 
finding  such  a  value  for  the  unknown  quantity,  as  will 
satisfy  the  equation  ;  that  is,  such  a  value  as,  being  sub- 
stituted for  the  unknown  quantity,  will  render  the  two  mem- 
bers equal.  This  is  called  a  ROOT  of  the  equation. 

A  Hoot  of  an  equation  is  said  to  be  verified,  when  being 
substituted  for  the  unknown  quantity  in  the  given  equation, 
the  two  members  are  found  equal  to  each  other. 

1.  Take  the  equation, 


Clearing  of  fractions  (Art.  104),  and  performing  the  opera- 
tions indicated,  we  have, 

12*  —  32   =   4x  —  8  +  24. 

107.  What  is  the  solution  of  an  equation?  What  is  the  found  value 
of  the  unknown  quantity  called  ?  When  is  a  root.  of  an  equation  said  to 
he  verified. 


SOLUTION      OF      EQUATIONS.  Ill 

Transposing  all  the  unknown  terms  to  the  first  member, 
and  the  known  terms  to  the  second  (Art.  106),  we  hare, 

12x  —  4x  —    —  8  +  24  +  32. 
Reducing  the  terms  in  the  two  members, 

8z   =   48. 

Dividing  both  members  by  the  coefficient  of  *, 

48 

:  T  : 

VERIFICATION. 

3X6  4(6  —  2)    , 

— 4   =  -  — i  +  3  ;    or, 

L  o 

+  9  —  4   =   2  +  3   =   5. 
Hence,  6  satisfies  the  equation,  and  therefore,  is  a  root. 

1O8.  By  processes  similar  to  the  above,  all  equations  of 
the  first  degree,  containing  but  one  unknown  quantity,  may 
be  solved. 

KULE. 

I.  Clear  the  equation  of  fractions,  and  perform  all  the 
indicated  operations : 

II.  Transpose  all  the  unknown  terms  to  the  first  member, 
and  all  the  known  terms  to  the  second  member : 

III.  ^Reduce  all  the  terms  in  the  first  member  to  a  single 
term,  one  factor  of  which  will  be  the  unknown  quantity, 
and  the  other  factor  will  be  the  algebraic  sum  of  its  coeffi- 
cients : 

IV.  Divide  both  members  by  the  coefficient  of  the  unknown 
quantity  :  the  second  member  will  then  be  the  value  of  the 
unknown  quantity. 

108.  Give  the  rule  for  solving  equations  of  the  first  degree  with  one 
unknown  quantity. 


112  ELEMENTARY      ALGEBRA. 

EXAMPLES. 

1.  Solve  the  equation, 

5x       4x  7       13a; 

13  = • 

12         3  86 

Clearing  of  fractions, 

lOx  —  32a;  —  312   =  21  —  52x. 

By  transposing, 

lOa  —  32x  +  52x  -    21  +  312. 
By  reducing,  30a  =  333 ; 

333         111 
hence,  x      :  —  =  —  =     11.1; 

a  result  which  may  be  verified  by  substituting  it  for  x  in 
the  given  equation. 

2.  Solve  the  equation, 

(3a  —  x)  (a  —  b)  +  2ax  =  4J(a  -f  a). 

Performing  the  indicated  operations,  we  have, 

3a2  —  ax  —  Sab  +  bx  +  2ax  =  4&c  +  4«5. 
By  transposing, 

—  ax  +  bx  +  2ax  —  4&B  =  4a5  +  Sab  —  3a*. 
By  reducing,  ax  —  3bx  =  *ldb  —  3a2 ; 

Factoring,  (a  —  3b)x  =  lab  —  3a2. 

Dividing  both  members  by  the  coefficient  of  x, 

lab  -  3a2 
a  —  3b 

3.  Given    3x  —  2  +  24  =  .31    to  find  x.     Ans.   x  =  3. 

4.  Given    x  +  18  =  3x  —  5   to  find  x.    Ans.  x=  11 J- 


SOLUTION      OF      EQUATIONS.  113 

6.  Given    6  —  2x  +  10   ;=  20  —  3x  —  2,  to  find  a. 

Ans.   x  =  2. 

6.  Given    x  +  IJB  -f  \x  =  11,  to  find  x.     Ans.  x  =  6. 

7.  Given    2x  —  %x  +  1   =  5x  —  2,  to  find  x. 

Ans.  x  =  %. 

Solve  tlie  following  equations: 

a  6  —  8a 

8.  Sax  -\  ---  3  =  Ix  —  a.         Ans.  x  =  --  =•• 

2  Qa  —  25 

x  —  3       x  x  —  19 

9.  —  —  +  -   =  20  --       -  •  Ans.   x  =  231  . 

•  0  .6 

x  +  3    ,    a;  x  —  5 

10.     --  +       =  4  --  -  --  ^W5.   x  =  3- 


x       3a;   .  4» 

11.    T  --  —  +  x  =  —  —  3.  Ans.  x  =  4. 

V  2  O 


12.    ---  ^  --  4  =  f.  Ans,.   x  =  —^  -  =- 

c          d  Sad  —  2bc 

x  —  a       2x  —  35       a  —  x 


. 

245. 


13. 


a        8—  a;       5  +  ajll 
14.    -  -  _  ---  —  +  -  =  0.      ^.   «  =  12. 

a  +  c   ,   a  —  c  252  a2  — 

I  >%  ^_^^^___^        I        _______       —       _ 

iw.  —          „  _ 

a  +  a       a  —  cc        a2  —  x2 

Sax  —  b       35  —  c 
16.       -^  ----  2—  =  4-5. 

56  +  95  — 

.  a;  =  -  --— 
16a 

x       x  —  2       x        13 
17     5  -~3-  +  2   =   T-  ^'   »= 


114  ELEMENTARY      ALGEBRA. 


?:         X    —    f 

c~  d~f' 
Ans.   x  = 


, 
oca  —  aca  +  aoa  —  abc 


NOTE.  —  What  is  the  numerical  value  of  «,  when   a  =  1, 
=  2,    c  =  3,    <?  —  4,    and  /  =  6  ? 


19.        -        - 


3x  —  5    ,    4x  —  2 
20.    a;  ----  (-•  --  =  a;  +  1.        J.ns.   x  =  6,  . 

13  11 


21.    a;  H  ---  1  ----  =   2«  —  43.             ^4^5.   a;  =  60. 
456 

4x  —  2  3a;  —  1                           . 

A.  ^  •      ^»C   ~~~   '                       —  —  ^^—  —  ^-^  •                                  jfjLftS*     *C    —    O» 

5  2 

,    bx  —  d  3a  +  d 

23.    3x  -J  ---  .  =  x  4-  «.            -4^5.   «  =  -  --  7-  • 

3  6  -t-  b 


ax  —  b       a        bx       bx  —  a 

24. —   = • 

4^32  3 


A  35 

Ans.  x  =  - — 


3a  —  26" 
4a5  20  -  4x  _     15  2 

JbO«  -  '  ~~~  '        -        -  *  -ii/(O«         »C       —  _        O  • 

5   —  JB  JB  X  11 

2a;  +  1        402  —  3x  471  —  6z 

20.    ----   —  9 


29  12  2 

Ans.   x  =  72. 

„*     fa  +  b}(x-b)  4ab-b*  a?-bx 

—^^r~  ~^+b~  ~T~ 

a4  +  3a35  +  4a~bz  —  Gab3  +  25* 
~  aft  -  52    " 


PROBLEMS.  115 

PROBLEMS. 

1O9.  A  PKOBLEM  is  a  question  proposed,  requiring  a 
solution. 

The  SOLUTION  of  a  problem  is  the  operation  of  finding  a 
quantity,  or  quantities,  that  will  satisfy  the  given  conditions. 

The  solution  of  a  problem  consists  of  two  parts  : 

I.  The  STATEMENT,  which  consists  in  expressing,  algebra- 
ically, the  relation  between  the  known  and  the  required 
quantities. 

IE.  The  SOLUTION,  which  consists  in  finding  the  values 
of  the  unknown  quantites,  in  terms  of  those  which  are 
Jcnoicn. 

The  statement  is  made  by  representing  the  unknown 
quantities  of  the  problem  by  some  of  the  final  letters  of  the 
alphabet,  and  then  operating  upon  these  so  as  to  comply 
with  the  conditions  of  the  problem.  The  method  of  stating 
problems  is  best  learned  by  practical  examples. 

1.  What  number  is  that  to  which  if  5  be  added,  the  sum 
will  be  equal  to  9  ? 
Denote  the  number  by  x.    Then,  by  the  conditions, 

x  +  5   =   9. 
This  is  the  statement  of  the  problem. 

To  find  the  value  of  x,  transpose  5  to  the  second  member ; 
then, 

x   =.   9  —  5  =  4. 

This  is  the  solution  of  the  equation. 

VERIFICATION. 

x  +  5  =  9. 

I'i9.  What  is  a  problem?  Wait  is  the  solution  of  a  problem?  Of 
how  many  parts  does  it  consist*  What  are  they?  What  is  the  state- 
niwit  ?  What  is  the  solution  ? 


116  E  L  E  M  E  N  T  A  K  V       A  L  G  E  B  U  A  . 

2.  Find  a  number  such  that  the  sum  of  one-half,  one-third, 
and  one-fourth  of  it,  augmented  by  45,  shall  be  equal  to  448 

Let  the  required  number  be  denoted  by       x. 

M 

Then,  one-half  of  it  trill  be  denoted  by       —  , 

2 

o* 

one-third  "  by       -, 

O 

B 

one-fourth      "  by       -; 

and,  by  the  conditions, 


This  is  the  statement  of  the  problem. 

Clearing  of  fractions, 

6x  +  4x  +  Sx  +  540  =  5376  , 
Transposing  and  collecting  the  unknown  terms, 
13x  =  4836; 

4836 
hence,  x  =  —  —   =  372. 

Id 

VERIFICATION. 

^*72         ^5*72         ^72 

-  +  —  -  +  -  -  +  45   =  186  +  124  +  93  +  45   =   448. 

±t  O  4 

3.  What  number  is  that  whose  third  part  exceeds  its 
fourth  by  16  ? 

Let  the  required  number  be  denoted  by  x.    Then, 

-x  —    the  third  part, 
3 

-x  =    the  fourth  part  ; 


P  It  ()  Ii  L  K  M  S  .  117 

and,  by  tlie  conditions  of  the  problem, 
\x  -  \x  =  16. 

Tbis  is  the  statement.     Clearing  of  fractions, 

4a;  —  3x  =   192, 
and  hence,  x  =  192. 

VERIFICATION. 


4.  Divide  $1000  between  A,  .Z?,  and   (7,  so  that  A  shall 
have  $72  more  than  JB,  and  C  $100  more  than  A. 

Let  x  denote  the  number  of  dollars  which  S  received. 

Then,  x  =  B's  number, 

x  +  72  =  A's  number, 
and,  a;+l72  =  C's  number; 

and  their  sum,  3x  +  244  —  1000,  the  number  of  dollars. 

This  is  the  statement.     By  transposing, 

Sx  =   1000  —  244   =   756  ; 
and,  x  =    •     -   =  252  =  IPs  share. 

Hence,    x  +     72     =  252  -f     72  =  324  =    A's  share, 
and,  x  +   172    =  252  +  172  =  424  =    <7's  share. 

VERIFICATION. 

252  +  324  -f  424   =   1000. 

5.  Out  of  a  cask  of  wine  which  had  leaked  away  a  third 
part,  21  gallons  were  afterwards  drawn,  and  the  cask  being 
then  guaged,  appea7%ed  to  be  half  full  :  how  much  did  it 
hold  ? 


118  ELEMENTARY       ALGEBRA. 

Let      x  denote  the  number  of  gallons. 

fly 

Then,  -    =    the  number  that  had  leaked  away. 
3 

and,          -  +  21    =    what  had  leaked  and  been  drawn. 
3 

x  x 

Hence,  by  the  conditions,   -  +  21    =    -  • 

3  2 

This  is  the  statement.     Clearing  of  fractions, 

2x  +  126   =  3x, 
and,  -  x  =    —  126  ; 

and  by  changing  the  signs  of  both  members,  which  does  not 
destroy  their  equality  (since  it  is  equivalent  to  multiplying 
both  members  Jby  —  1),  we  have, 

x  =  126. 

VERIFICATION. 

-  +  21    =   42  +  21    =   63   i=  - 


6.  A  fish  was  caught  whose  tail  weighed  9  Ibs.,  his  head 
weighed  as  much  as  his  tail  and  half  his  body,  and  his  body 
weighed  as  much  as  his  head  and  tail  together :  what  was 
the  weight  of  the  fish  ? 

Let  2x  =  the  weight  of  the  body,  in  pounds. 

Then,      9  +  x  =  weight  of  the  head ; 

and  since  the  body  weighed  as  much  as  both  head  and  tail, 

2x  =  9  +  9  -f  x, 
which  is  the  statement.    Then, 

2x  —  x  =   18,   and   x  =   18. 


PROBLEMS.  119 

Hence,  we  have, 

2x  =  36$.  =  weight  of  the  body, 

9  +  x  =  27$.  —  weight  of  the  head, 

9$.  =  weight  of  the  tail ; 

hence,  72$.  =  weight  of  the  fish. 

7.  The  sum  of  two  numbers  is  67,  and  their  differer  ce  19 
what  are  the  two  numbers  ? 

Let      x  denote  the  less  number. 

Then,  x  -f-  1 9  =  the  greater ;  and,  by  the  conditions, 

2x  +  19  =  67. 
This  is  the  statement.    Transposing, 

2x  =  67  —  19  =  48; 

hence,  x  =  —  =  24,   and  x  +  19  =  43. 

VERIFICATION. 

43  +  24  =   67,   and   43  —  24  =   19. 

ANOTHER    SOLUTION. 

Let  x  denote  the  greater  number. 

Then,        x  —  19   will  represent  the  less, 
and,  2x  —  19    =    67  ;    whence    2ce  =  67  -f  19. 

o/> 

Therefore,  x  =  -  -  :_  43 ; 

2i 

and,  consequently,    x  —  19  =  43   -  19  =  24. 

GENERAL    SOLUTION    OF    THIS    PROBLEM. 

The  sum  of  two  numbers  is  s,    their  difference  ia  d:  what 
are  the  tAVO  numbers  ? 


120  ELEMENTARY      ALGEBRA.     ~ 

Let  x  denote  the  less  number. 

Then,        a;  +  d  will  denote  the  greater, 
and  2x  +  d   =  s,  their  sum.    Whence, 

s  —  d        s       d 
~2~      =  2~2*' 
and,  consequently, 

s       d       ,        s       d 
«+*=j-5+*=5+V 

As  these  two  results  are  not  dependent  on  particular 
values  attributed  to  s  or  d,  it  follows  that : 

1.  The  greater  of  two  numbers  is  equal  to  half  their  sum, 
plus  half  their  difference  : 

2.  The  less  is  equal  to  half  their  sum,  minus  half  their 
difference. 

Thus,  if  the  sum  of  two  numbers  is  32,  and  their  differ- 
ence 16, 

32        16 
the  greater  is,        —  +  --  ==16  +  8  =  24  ;    and 

2t  2i 

32       16 
the  less,  — —  =16  —  8=     8. 

2i  2i 

VERIFICATION. 

24  +  8  =  32;   and   24  —  8  =  16. 

8.  A  jv*rson  engaged  a  workman  for  48  days.  For  each 
day  that  he  labored  he  received  24  cents,  and  for  each  day 
that  he  vras  idle,  he  paid  12  cents  for  his  board.  At  the 
end  of  the  48  days,  the  account  was  settled,  when  the  laborer 
received  504  cents.  Required,  the  number  of  working  days, 
and  the  number  of  days  he  was  idle. 

If  the  number  of  working  days,  and  the  number  of  idle 
days,  were  known,  and  the  first  multiplied  by  24,  and  the 


r  u  o  B  L  E  M  s .  121 

second  by  12,  the  difference  of  these  products  would»be. 
504.  Let  us  indicate  these  operations  by  means  of  algebraic 
signs. 

Let  x  denote  the  number  of  working  days. 

Then,      48  —  x    =   the  number  of  idle  days, 

24  x  x   =    the  amount  earned, 
and,       12(48  —  x}  =    the  amount  paid  for  board. 

Then,  24a  -  12(48  —  a;)   =  504, 

what  was  received,  which  is  the  statement. 
Then,  performing  the  operations  indicated, 

24*  —  576  +  12as  =     504, 
or,  3Qx  —  504  +  570   =   1080, 

and,  x  =  -     -  =  30,  the  number  of  working  days ; 

3G 

whence,       48  —     30     =18,  the  number  of  idle  days. 

VERIFICATION. 

Thirty  days'  labor,  at  24  cents  )  n 

J      '  1 30  X  24  =  720  cents, 

a  day,  amounts  to ) 

And  18  days'  board,  at  12  cents  ) 

J                                    }•  18  X  12  =  216  cents, 
a  day,  amounts  to )  • 

The  difference  is  the  amount  received  .    504  cents. 


GENERAL    SOLUTION. 

*  This  problem  may  be  made  general,  by  denoting  the  whole 
number  of  working  and  idle  days,  by  n ; 

The  amount  received  for  each  day's  work,  by  a ; 

The  amount  paid  for  board,  for  each  idle  day,  by  b ; 

And  what  was  due  the  laborer,  or  the  balance  of  the 
account,  by  c. 
6 


122  K  L  E  M  K  N  T  A  R  Y       ALGEBRA. 

^ft  before,  let  the  number  of  working  days  be  denoted 
by  x. 

The  number  of  idle  days  will  then  be  denoted  by  n  —  x. 

Hence,  what  is  earned  will  be  expressed  by  ax,  and-  the 
sura  to  be  deducted,  on  account  of  board,  by  b(n  —  x). 

The  statement  of  the  problem,  therefore,  is, 

ax  —  b(n  —  x)   =  c. 
Performing  indicated  operations, 

ax  —  bn  -f  bx  —  c,    or,    (a  4-  f>}x  =  c  -f-  bn ; 

whence,  x  =   -      -—    —     number  of  working  days ; 

(Ju    ~*|~     (s 

c  +  bn       an-\-bn—c—bn 

and,  n  —  x  =  n —j-  =  -        —— v , 

a+  b  .«  +  o 

or,  n  —  x  =  —  =    number  of  idle  davs. 

a  +  b 

Let  us  suppose  n  =  48,  a  —  24,  b  =  12,  and  c  —  504  ; 
these  numbers  will  give  for  x  the  same  value  as  before 
found. 

9.  A  person  dying  leaves  half  of  his  property  to  his  wife, 
one-sixth  to  each  of  two  daughters,  one-twelfth  to  a  servant, 
and  the  remaining  $600  to  the  poor ;  what  was  the  amount 
of  the  property  ? 

Let          x  denote  the  amount,  in  dollars, 

M 

Then,       ^   —      what  he  left  to  his  wife, 

2t 

JM 

-    =      what  he  left  to  one  daughter,  * 

O/k*  /£ 

and,  —  =  -  what  he  left  to  both  daughters, 

u  3 

X 
also,  —  =       what  he  left  to  his  servant, 

and,        $600  =       what  he  left  to  the  poor. 


PROBLEMS.  123 

Then,  by  the  conditions, 

/j«  /•«  />• 

-  +  -  4-  —  +  600  —  x,   the  amount  of  the  property, 

— .  t  >  1   w 

which  gives,  x  =  $7200. 

10.  A  and  J3  play  together  at  cards.    A  sits  down  with 
$84,  and  _Z>  with  $48.     Each  loses  and  wins  in  turn,  wrhen 
it  appears  that  A  has  five  times  as  much  as  J5.    How  much 
did  A  win  ? 

Let  x  denote  the  number  of  dollars  A  won. 
Then,  A   rose  with   84  +  x  dollars, 

and  J?  rose  with   48  —  x  dollars. 

But,  by  the  conditions,  we  have, 

84  +  x  =  5(48  —  aj), 
hence,  84  +  x  —   240  —  5x; 

and,  Qx  =   156, 

consequently,  x  =  26  ;   or  A  won  $26. 

VERIFICATION. 

84  -f  26   =  110  ;     48  —  26  =  22; 
110   =   5(22)    =   110. 

11.  A  can  do  a  piece  of  work  alone  in  10  days,  JB  in  15 
days ;  in  what  time  can  they  do  it  if  they  work  together  ? 

Denote  the  time  by  &,   and  the  work  to  be  done,  l»y  1. 
Then,  in 

1  day,  A  can  do  —  of  the  work,  and 

JB  can  do  —  of  the  work ;  and  in 
13 

•C 
x  days,  A  can  do  —    of  the  work,  and 

•C 

B  can  do   ~  of  the  work. 
13 


124  K  L  E  M  E  N  T  A  14  Y      ALGEBRA. 

Hence,  by  the  conditions, 

•C  *C 

•- (-  —  =   1,  which  gives,    13x  +  Wx  —   130; 

10    13 

130 

hence,    23x  =130,  x  =  —  =  5£f  days. 

23 

12.  A  fox,  pursued  by  a  hound,  has  a  start  of  60  of  his 
own  leaps.  Three  leaps  of  the  hound  are  equivalent  to  7  of 
the  fox  ;  but  while  the  hound  makes  6  leaps,  the  fox  makes 
9 :  how  many  leaps  must  the  hound  make  to  overtake  the 
fox? 

There  is  some  difficulty  in  this  problem,  arising  from  the 
different  units  which  enter  into  it. 

Since  3  leaps  of  the  hound  are  equal  to  7  leaps  of  the  fox, 

7 

1  leap  of  the  hound  is  equal  to  -  fox  leaps. 

3 

Since,  while  the  hound  makes  6  leaps,  the  fox  makes  9, 

9  3 

while  the  hound  makes  1  leap,  the  fox  will  make  - ,  or  - 

0  — 

leaps. 

Let  x  denote  the  number  of  leaps  which  the  hound  makes 
before  he  overtakes  the  fox ;  and  let  1  fox  leap  denote  the 

unit  of  distance. 

7 
Since  1  leap  of  the  hound  is  equal  to   -   of  a  fox  leap,  x 

7 
leaps  will  be  equal  to  -x  fox  leaps ;  and  this  will  denote  the 

0 

distance  passed  over  by  the  hound,  in  fox  leaps. 

'  3 

Since,  while  the  hound  makes  1  leap,  the  fox  makes  ^ 

3 
leaps,  while  the  hound  makes  x  leaps,  the  fox  makes  -x  leaps ; 

m 

and    this    added    to    60,    his    distance    ahead,   Avill    give 

g 

-x  +  CO,    for  the  whole  distance  passed  over  by  the  fox. 

I 


PROBLEMS.  125 


Hence,  from  the  conditions, 


7  3 

-a;  =  -x  +  60 ;  whence, 

Ux  =  Qx  +  360; 

x  =  72. 


The  hound,  therefore,  makes  72  leaps  before  overtaking 
le  fo 

leaps. 


Q 

the  fox;  in  the  same  time,  the  fox  makes   72  x  -  =  108 

m 


VERIFICATION. 

108  +  60  =:  168,    whole  number  of  fox  leaps, 
72  X     I  -  168. 

o 

13.  A  father  leaves  his  property,  amounting  to  $2520,  to 
four  sons,  A,  _Z?,  (7,  and  D.     C  is  to  have  $360,  J3  as  much 
as  C  and  D  together,  and  A  twice  as  much  as  .#,  less  $1000 : 
how  much  do  A,  B-,  and  D  receive  ? 

Am.   A,  $760;  .#,  $880;  Z>,  $520. 

14.  An  estate  of  $7500  is  to  be  divided  among  a  widow, 
two  sons,  and  three  daughters,  so  that  each  son  shall  receive 
twice  as  much  as  each  daughter,  and  the  widow  herself  $500 
more  than  all  the  children :  what  was  her  share,  and  what 
the  share  of  each  child  ? 

{Widow's  share,     $4000. 
Each  son's,  1000. 

Each  daughter's,       500. 

15.  A  company  of  180  persons  consists  of  men,  women, 
and  children.     The  men  are  8  more  in  number  than  the 
women,  and  the  children  20  more  than  the  men  and  women 
together :  how  many  of  each  sort  in  the  company  ? 

Ans.   44  men,  36  women,  100  children. 


126  E  L  E  M  E  N  T  A  II  Y       A  L  „  E  BRA. 

16.  A  father  divides  $2000  among  five  sons,  so  that  each 
elder  should  receive  $40  more  than  his  next  younger  bro- 
ther :  what  is  the  share  of  the  youngest?  Ans.   $320. 

17.  A  purse  of  $2850  is  to  be  divided  among  three  per- 
sons, A,  J3,  and  C.    A's  share  is  to  be  to  jB's  as  6  to  1 1 , 
and  C  is  to  have  $300  more  than  A  and  B  together :  what 
is  each  one's  share?  A>s,  $450  ;  B's,  8825  ;  (7's,  $1575. 

18.  T\vo  pedestrians  start  from  the  same  point  and  travel 
in  the  same  direction ;    the  first  steps  twice  as  far  as  the 
second,  but  the  second  makes  5  steps  while  the  first  makes 
but  one.     At  the  end  of  a  certain  time  they  are  300  feet 
apart.     Now,  allowing  each  of  the  longer  paces  to  be  3  feet, 
how  far  will  each  have  traveled  ? 

Ans.    1st,  200  feet ;  2d,  500. 

19.  Two  carpenters,  24  journeymen,  and  8  apprentices 
received  at  the  end  of  a  certain  time  $144.     The  carpenters 
received  $1  per  day,  each  journeyman,  half  a  dollar,  and 
each  apprentice,  25  cents :  how  many  days  were  they  em- 
ployed? Ans.    9  days. 

20.  A  capitalist  receives  a  yearly  income  of  $2940  ;  four- 
fifths  of  his  money  bears  an  interest  of  4  per  cent.,  and  the 
remainder  of  5  per  cent. :  how  much  has  he  at  interest  ? 

Ans.   $70000. 

21.  A  cistern  containing  60  gallons  of  water  has  three 
unequal  cocks  for  discharging  it ;  the  largest  will  empty  it 
in  one>hour,  the  second  in  two  hours,  and  the  third,  in  three: 
in  what  time  will  the  cistern  be  ehipticd  if  they  all  run  to- 
gether ?  Ans.    32T8T  min. 

22.  In  a  certain  orchard,  one-half  are  apple  trees,  one- 
fourth  peach  trees,  one-sixth  plum  trees;  there  are  also,  120 
cherry  trees,  and  80  pear  trees :    how  many  trees  in  the 
orchard?  Ans.  2400. 

23.  A   farmer   being   asked   ho\\    many   sheep   ho   had, 


P  ii   )  B  L  K  M  8  .  127 

answered,  that  he  had  them  in  five  fields  ;  in  the  1st  he  had 
|,  in  the  2d,  £,  in  the  3d,  |,  and  in  the  4th,  TV,  and  in  the 
5th,  450  :  how  many  had  he  ?  Ans.  1200. 

24.  My  horse  and  saddle  together  are  worth  $132,  and 
the  horse  is  worth  ten  times  as  much  as  the  saddle :  what 
is  the  value  of  the  horse  ?  Ans.   $120. 

25.  The  rent  of  an  estate  is  this  year  8  per  cent,  greater 
than  it  was  last.    This  year  it  is  $1890:  what  was  it  last 
year?  Ans.   $1750. 

26.  What  number  is  that,  from  which  if  5  be  subtracted, 
|  of  the  remainder  will  be  40  ?  Ans.   65. 

27.  A  post  is  |  in  the  mud,  £  in  the  water,  and  10  feet 
above  the  water :  Avhat  is  the  whole  length  of  the  post  ? 

Ans.   24  feet. 

28.  After  paying  |  and  i  of  my  money,  I  had  66  guineas 
left  in  my  purse  :  how  many  guineas  were  in  it  at  first  ? 

Ans.   120. 

29.  A  person  was  desirous  of  giving  3  pence  apiece  to 
some  beggars,  but  found  he  had  not  money  enough  in  his 
pocket  by  8  pence;  he  therefore  gave  them  each  2  pence 
and  had  3  pence  remaining :  required  the  number  of  beg- 
gars. Ans.   11. 

30.  A  person,  in  play,  lost  %  of  his  money,  and  then  won 
3  shillings ;  after  which  he  lost  £  of  what  he  then  had ;  and 
this  done,  found  that  he  had  but  12  shillings  remaining: 
what  had  he  at  first  ?  Ans.  20s. 

31.  Two  persons,  A  and  J5,  lay  out  equal  sums  of  money 
in  trade;  A  gains  $126,  and  B  loses  $87,  and  A's  money  is 
then  double  of  B's  :  what  did  each  lay  out?        Ans.   $300. 

32.  A  person  goes  to  a  tavern  with  a  certain  sum  of 
money  in  his  pocket,  where  he  spends  2  shillings :  he  then 
borrows  as  much  money  as  he  had  left,  and  going  to  another 
tavern,  he  there  spends  2  shillings  also;   then  borrowing 


128  ELK  MEN  TAUT      ALGEBRA. 

again  as  much  money  as  was  left,  he  went  to  a  third  tavern, 
where  likewise  he  spent  2  shillings,  and  borrowed  as  much 
as  he  had  left :  and  again  spending  2  shillings  at  a  fourth 
tavern,  he  then  had  nothing  remaining.  What  had  he  at 
first  ?  Ans.  3s.  9 d. 

33.  A  tailor  cut  19  yards  from  each  of  three  equal  pieces 
of  cloth,  and  17  yards  from  another  of  the  same  length, 
and  found  that  the  four  remnants  were  together  equal  to 
142  yards.     How  many  yards  in  each  piece ?  Ans     54. 

34.  A  fortress  is  garrisoned  by  2600  men,  consisting  of 
infantry,  artillery,  and  cavalry.     Now,  there  are  nine  times 
as  many  infantry,  and  three  times  as  many  artillery  soldiers 
as  there  are  cavalry.     How  many  are  there  of  each  corps  ? 

Ans.    200  cavalry;  COO  artillery  ;  1800  infantry. 

35.  All  the  journeyings  of  an  individual  amounted  to  2970 
miles.     Of  these  he  traveled  3|  times  as  many  by  water  as 
on  horseback,  and  2|  times  as  many  on  foot  as  by  water. 
How  many  miles  did  he  travel  in  each  way  ? 

Ans.    240  miles;  840  m. ;  1890  m. 

36.  A  sum  of  money  was  divided  between  two  persons, 
A  and  J5.     A's  share  was  to  J?'s  in  the  proportion  of  5  to  3, 
and  exceeded  five-ninths  of  the  entire  sum  by  50.     "What 
was  the  share  of  each?      Ans.   A's  share,    450;  JFs,    270. 

37.  Divide  a  number  «  into  three  such  parts  that  the 
second  shall  be  n  times  the  first,  and  the  third  m  times  as 
great  as  the  first. 

a  na  ma 

2d»  T-T-;  rrrr,  '•>  3d> 


±  OC«  «  ^*U»*  .5          wi« 

1  +  m  +  n  1  +  m  +  n  1  +  m  +  n 

38.  A  father  directs  that  $1170  shall  be  divided  among 

his  three  sons,  in  proportion  to  their  ages.     The  oldest  ia 

t\vice  as  old  as  the  youngest,  and  the  second  is  one-third 

older  than  the  youngest.     How  much  was  each  to  receive? 

A/tv.    $'270,  youngest  ;  *:!UO,  second  ;  $540,  <  Ulest. 


PROBLEMS.  129 

39.  Three  regiments  are  to  furnish  594  men,  and  each  to 
furnish  in  proportion  to  its  strength.     Now,  the  strength  of 
the  first  is  to  the  second  as  3  to  5  ;  and  that  of  the  second 
to  the  third  as  8  to  7.     How  many  must  each  furnish  ? 

Ans.    1st,  144  men  ;  2d,  240  ;  3d,  210. 

40.  Five  heirs,  A,  J?,  (7,  D,  and  E,  are  to  divide  an  inher- 
itance of  $5600.     -B  is  to  receive  twice  as  much  as  A,  and 
8200  more ;   C  three  times  as  much  as  A,  less  $400 ;  D  the 
half  of  what  -Z>  and  C  receive  together,  and  150  more ;  and 
E  the  fourth  part  of  what  the  four  others  get,  plus  $475. 
How  much  did  each  receive  ? 

A>s,  $500;  .#'s,  1200;   C"a,  1100;  D's,  1300;  E's,  1500. 

41.  A  person  has  four  casks,  the  second  of  which  being 
filled  from  the  first,  leaves  the  first  four-sevenths  full.     The 
third  being  filled  from  the  second,  leaves  it  one-fourth  full, 
and  when  the  third  is  emptied  into  the  fourth,  it  is  found  to 
fill  only  nine-sixteenths  of  it.     But  the  first  Avill  fill  the  third 
and  fourth,  and  leave   15  quarts  remaining.      How  many 
gallons  does  each  hold  ? 

Ans.    1st,  35  gal. ;  2d,  15  gal. ;  3d,  11}  gal. ;  4th,  20  gal. 

42.  A  courier  having  started  from  a  place,  is  pursued  by 
a  second  after   the  lapse  of  10  days.     The  first  travels  4 
miles  a  day,  the   other  9.      How  many  days  before  the 
second  will  overtake  the  first  ?  Ans.    8. 

43.  A  courier  goes  3H  miles  every  five  hours,  and  is  fol- 
lowed by  another  after  he  had  been  gone  eight  hours.     The 
second  travels  221  miles  every  three  hoxirs.     How  many 
hours  before  he  will  overtake  the  first  ?  Ans.   42. 

44.  Two  places  are  eighty  miles  apart,  and  a  person  leaves 
one  of  them  and  travels  towards  the  other  at  the  rate  of  Si- 
miles per  hour.     Eight  hours  after,  a  person  departs  from 

6* 


130  ELEMENTARY       ALGEBRA. 

the  second  place,  and  travels  at  the  rate  of  5|  miles  per  hour. 
How  long  before  they  will  be  together? 

A.ns.    6  hours. 

EQUATIONS    CONTAINING   TWO    UNKNOWN   QUANTITIES. 

11O.     If  we  have  a  single  equation,  as, 
2x  +  3y  =  21, 

containing  two  unknown  quantities,  %  and  y,  we  may  find 
the  value  of  one  of  them  in  terms  of  the  other,  as, 

21  -  3 


x  = 


(1.) 


Now,  if  the  value  of  y  is  unknown,  that  of  x  will  also  be 
unknown.  Hence,  from  a  single  equation,  containing  two 
unknown  quantities,  the  value  of  x  cannot  be  determined. 

If  we  have  a  second  equation,  as, 

5x  +  4y  =   35, 

we  may,  as  before,  find  the  value  of  x  in  terms  of  y,  giving, 

35  —  4y 

"--— (2° 

Now,  if  the  values  of  x  and  y  are  the  same  .in  Equations 
( 1 )  and  ( 2  ),  the  second  members  may  be  placed  equal  to 
each  other,  giving, 

21  —  3y         35  —  4y 

t  = ^-2  ,    or    105  -  15y  =   70  -  Sy ; 

£.  O 

from  which  we  find,  y  =  5. 

110.  In  one  equation  containing  two  unknown  quantities,  can  you  find 
the  value  of  either  ?  I£  you  hare  a  second  equation  involving  the  same 
two  unknown  quantities,  can  YOU  find  their  values?  What  are  such  equa- 
tions called  ? 


ELIMINATION.  131 

Subtituting  this  value  for  y  in  Equations  (1)  or  (2),  we 
6iid  x  =  3.  Such  equations  are  called  Simultaneous 
equations.  Hence, 

111.  SIMULTANEOUS  EQUATIONS  are  those  in  which  the 
Talues  of  the  unknown  quantity  are  the  same  in  both. 

ELIMINATION. 

112.  ELIMINATION  is  the  operation  of  combining  two 
equations,  containing  tAvo  unknown  quantities,  and  deducing 
therefrom  a  single  equation,  containing  but  one. 

There  are  three  principal  methods  of  elimination  : 

1st.  By  addition  or  subtraction. 

2d.   By  substitution. 

3d.   By  comparison. 
We  shall  consider  these  methods  separately. 

Elimination  by  Addition  or  Subtraction. 
1.  Take  the  two  equations, 

3x  —  2y  =     7, 
8x  -f-  2y  =  48. 

If  we  add  these  two  equations,  member  to  member,  we 

obtain, 

llaj  =  55; 

which  gives,  by  dividing  by  11, 

x  =  5; 

and  substituting  this  value  in  either  of  the  given  equations, 
we  find, 

y  =  4. 

111.  What  are  simultaneous  equations? 

112.  What  is  elimination?      How  many  methods  of  elimination  are 
there  ?    What  are  they  ? 


132  ELEMENTARY       ALGEBRA. 

2.  Again,  take  the  equations, 

Sx  +  2y  =  48, 
3z  +  2y   =   23. 

If  we  subtract  the  2d  equation  from  the  1st,  we  obtain, 

5x  =.  25; 
which  gives,  by  dividing  by  5, 

x  —   5; 

and  by  substituting  this  value,  we  find, 
V  =  4. 

3.  Given  the  sum  ot  two  numbers  equal  to  s,  and  theif 
difference  equal  to  <7,  to  find  the  numbers. 

Let  x  =  the  greater,  and  y  the  less  number. 

Then,  by  the  conditions, x  +  y  =  g. 

and,  x  —  y  =  d. 

By  adding  (Ait.  102,  Ax.  1), 2x  =  s  +  d. 

By  subtracting  (Art,  102,  Ax.  2),  .  .  .  2y  =  *  —  d. 

Each  of  these  equations  contains  but  one  unknown  quantity. 


From  the  first,  we  obtain, x  =  S- 

o         ' 


and  from  the  second, y  — 


These  are  the  same  values  as  were  found  in  Prob.  7,  page 
120. 

4.  A  person  engaged  a  workman  for  48  days.  For  each 
day  that  he  labored  he  was  to  receive  24  cents,  and  for  each 
day  that  he  was  idle  he  was  to  pay  1 2  cents  for  his  board. 
At  the  end  of  the  48  days  the  account  was  settled,  when  the 
laborer  received  504  cents.  Required  the  number  of  work- 
ing days,  and  the  number  of  days  he  was  idle. 


ELIMINATION.  133 

Let  a;  =     the  number  of  working  days, 

y  —     the  number  of  idle  days. 

Then.  24a  =     what  he  earned, 

and,  12y  =     what  he  paid  for  his  board. 

Then,  by  the  conditions  of  the  question,  we  have, 

x  +      y  =     48, 
and,  24«  —  12y  =  504. 

This  is  the  statement  of  the  problem. 

It  has  already  been  sho~wn  (Art.  102,  Ax.  3),  that  the  two 
members  of  an  equation  may  be  multiplied  by  the  same  num- 
ber, without  destroying  the  equality.  Let,  then,  the  first 
equation  be  multiplied  by  24,  the  coefficient  of  x  in  the 
second ;  we  shall  then  have, 

24<e  +  24y  =   1152 
24jc  —  12y  =     504 

and  by  subtracting,  36y  =     648 

•1 '"'''       '        •'•  "  =  i?  =  18-      ' 

Substituting  this  value  of  y  in  the  equation, 
24x  —  12y  =  504,    we  have,     24*  —  216  =  504; 

which  gives, 

720 
24a?  —  504  +  216  =  720,     and     x  =  — -  =  30. 

VERIFICATION'. 

x  +      y  =     48     gives  30  +  18  =     48, 

24«  —  12y  =  504     gives     24  X  80  —  12  X  18   =  504. 


134  ELEMENTARY       ALGEBRA. 

113.     In  a  similar  manner,  either  unknown  quantity  may 
be  eliminated  from  either  equation ;  hence,  the  folio Aving 


RULE. 


I.  Prepare  the  equations  so  that  the  coefficients  of  the 
quantity  to  be  eliminated  shall  be  numerically  equal: 

II.  If  the  signs  are  unlike,  add  the  equations,  member 
to  member  /  if  alike,  subtract  them,  member  from  member. 


EXAMPLES. 


Find  the  values  of  x  and  y,  by  addition  or  subtraction, 
in  the  following  simultaneous  equations  : 


.,    j  4x  -  >ly  =    —  22  ) 

6.  •{  Ans.  x  =  5,  y  =  6. 
(  5x  +  2y  =        37  ) 

„    j  2x  +  Gy  =  42  ) 

7.  -j  Ans.   x  =  44,   y  =  54. 
(  8x  —  Qy  =     3  ) 

j  8x  —  9y  =   1    ) 

8.  i  Ans.  x  —  4,  y  —  i. 
(  6cc  —  3  =  4z  ) 


.  J  a;  =  6,  y  =  9. 


11.  X  7  V  4ns.  <  x  =  14,  y  =  16. 


113.  What  is  the  rule  fa?  elimination  by  addition  or  subtraction? 


ELIMINATION.  135 

12.  Says  A  to  .Z?,  you  give  me  $40  of  jour  money,  and 
I  shall  then  have  five  times  as  much  as  you  will  have  left. 
Xow  they  both  had  $120  :  how  much  had  each? 

Ans.  Each  had  $60. 

13.  A  father  says  to  his  son,  "  twenty  years  ago,  my  age 
was  four  times  yours  ;  now  it  is  just  double  :  "  what  were 

their  ages  ?  /  A       i  Father's,  60  years. 

Ans.  S  ,., 

(  Son's,       30  years. 

• 

14.  A  father  divided  his  property  between  his  two  sons. 
At  the  end  of  the  first  year  the  elder  had  spent  one-quarter 
of  his,  and  the  younger  had  made  $1000,  and  their  property 
was  then  equal.     After  this  the  elder  spent  $500,  and  the 
younger  made  $2000,  when  it  appeared  that  the  younger  had 
just  double  the  elder:  what  had  each  from  the  father? 

.       j  Elder,        $4000. 
'  (  Younger,  $2000. 

15.  If  John  give  Charles  15  apples,  they  will  have  the 
same  number;  but  if  Charles  give  15  to  John,  John  will 
have  15  times  as  many,  wanting  10,  as  Charles  will  have  left. 
How  many  has  each  ?  j       j  J°^n>       50. 

S-  1  Charles,  20. 

16.  Two  clerks,  A  and  _Z?,  have  salaries  which  are  together 
equal  to  $900.     A  spends  T^  per  year  of  what  he  receives, 
and  _B  adds  as  much  to  his  as  A  spends.    At  the  end  of  the 
year  they  have  equal  sums:  what  was  the  salary  of  each? 

A>s  =  $500. 


Elimination  by  Substitution. 
114.     Let  us  again  take  the  equations, 

Sa;  +  ty  =  43,  (  1.) 

Ux  +  9y  =  69.  (2.) 

1  1  4.  Give  the  rule  for  eliminaticm  by  substitution.    When  is  this  method 
t>sed  to  the  greatest  advantage? 


136  ELEMENTAKY      ALGEBRA. 

Find  the  value  of  x  in  the  first  equation,  which  gives, 


Substitute  this  value  of  x  in  the  second  equation,  and  w« 
have, 

11  x  -L_J£+  gy  =  69; 


or,,  473  —  77y  -f  45y  =  345  ; 

or,  —  32y  =  —  128. 

Here,   ar  has  been  eliminated  by  substitution. 

In  a  similar  manner,  we  can  eliminate  any  unknown  quan- 
tity ;  hence,  the 

RULE. 

I.  Find  from  either  equation  the  value  of  the  unknown 
quantity  to  be  eliminated: 

II.  Substitute  this  value  for  that  quantity  in  the  other 
equation. 

NOTE.  —This  method  of  elimination  is  used  to  great  advan- 
tage when  the  coefficient  of  either  of  the  unknown  quantities 
is  1. 

EXAMPLES. 

Find,  by  the  last  method,  the  values  of  x  and  y  in  the 
following  equations  : 

1.  3cc  —  y  =  1,    and    3y  —  2x  —  4. 

Ans.   x  =   1,    y  =  2. 

2.  5y  —  4x  =   —  22,    and    3y  +  4x  =  38. 

Ans.   x  =  8,    y  •=.  2. 

3.  x  +  8y  =  18,    and    y  —  Sx  =   —  29. 

Ans.  x  =  10,    y  =  1. 


ELIMINATION.  137 


2 

4.    5x  -•  y  =   13,    and    Sx  -f  -y  =   29. 

y 


Ans.   x  =  3£,    y  = 


5.    103  —       =   69,    and    lOy  —     " '=  49. 
5  7 


6.    x  +    *  -       =  10,    and       +        =  2. 


x  =  8,    y  =   10. 

7.    |  -  |  +  5   =   2,    a;  +  |  =   I7f 

Ans.   x  =   15,    y  =   14. 

8-    1  +  5  +  3   =   8*.    and    7~*  =  I' 

•.  «  =  3|,    y  =  4. 

y 


n         _       i    «   .  .   K 

8        4  H  5)  12        16 

Ans.  x  =  12,    y  =  16. 

10.  |  -  y  -  1   =    -  9,    and    5a:  -  ^  =  29. 

^Iws.  a;  =  6,    y  =  7. 

11.  Two  misers,  -4  and  jB,  sit  down  to  count  over  their 
money.     They  both  have  $20000,  and  B  has  three  times  as 
much  as  A  :  how  much  has  each  ? 


,   A 

A  \  ^3., 

'    (^, 


$15000. 


12.  A  person  has  two  purses.  If  he  puts  87  into  the  first, 
the  whole  is  worth  three  times  as  much  as  the  second  purse  : 
but  if  he  puts  $7  into  the  second,  the  whole  is  worth  five 
times  as  much  as  the  first:  what  is  the  value  of  each  purse? 

Ans.    1st,  $2  ;  2d,  $3. 


138  K  L  E  M  K  N  T  A  Ii  Y       A  L  G  K  B  K  A  . 

13.  Two  numbers  have  the  following  properties  :  if  the 
first  be  multiplied  by  6,  the  product  will  be  equal  to  the 
second  multiplied  by  5  ;    and   1  subtracted  from  the  first 
leaves  the  same  remainder  as  2  subtracted  from  the  second  : 
what  are  the  numbers  ?  Ans.   5  and  6. 

14.  Find  two  numbers  with  the  following  properties  :  the 
first  increased  by  2  is  3£  times  as  great  as  the  second  ; 
and  the  second  increased  by  4  gives  a  number  equal  to  half 
the  first  :  what  are  the  numbers  ?  Ans.    24  and  8. 

15.  A  father  says  to  his  son,  "twelve  years  ago,  I  was 
twice  as  old  as  you  are  now:  four  times  your  age  at  that 
time,  plus  twelve  years,  will  express  my  age  twelve  years 
hence  :  "    what  were  their  ages  ? 

(  Father,  72  years. 
AnS'    \  Son,        30       « 

Elimination  by   Comparison. 

115.    Take  the  same  equations, 

5x  +  1y  =  43 
llx  +  Qy  =  69. 

Finding  the  value  of  x  from  the  first  equation,  we  have, 
43  —  iy 

iyt       _        __  5_    • 

~T~ 

and  finding  the  value  of  x  from  the  second,  we  obtain, 

69  —  9y 


115.  Give  the  rule  for  climinatnn  by  comparison. 


ELIMINATION.  139 

Let  these  two  values  of  x  be  placed  equal  to  each  other, 
and  we  have, 

43  —  7y  __  69  —  9y 
5  11 

Or,  473  —  77y  =  345  -  45y; 

or,  —  32y  =   —  128. 

Hence,  y  —  4. 

69  —  36 

And,  x  =  — — — =  3. 

This  method  of  elimination  is  called  the  method  by  com- 
parison, for  which  we  have  the  following 

BULE. 

I.  Find,  from  each  equation,   the  value  of  the  same 
unknown  quantity  to  be  eliminated: 

II.  Place  these  values  equal  to  each  other. 

EXAMPLES. 

Find,  by  the  last  rule,  the  values  of  x  and  y,  from  the 
following  equations, 

1.  3a5  4-  ^  -f  6   =  42,    and    y  —  ^-  =   14£- 

Ans.   35  =   11,    y  =  15. 

2.  |  -  |  +  5   =  6,    and    |  -f  4  =  2-  +  6. 

Ans.  x  —  28,    y  =  20. 
•>        «c       22 

3'  15  -  4  +  T  =  ''  and  3y  ~  *  =  6' 

,4ns.  35  =  9,    y  =  5. 

4.    y  —  3  =  -a  +  5,    and    — ~=   =  y  —  3£. 

A  ^ 

^lw*.   a;  =  2,    y  =  9, 


14:0         ELEMENTARY   ALGEBRA. 

y-x   x  y_ 


Ans.  x  =  16,    y  =  7. 

2y 
=  as  -  -|,    and    s  +  y  =  H. 

.   a;  =   10,    y  =  6. 


—  — 

7.    -  --S  =  a>  -  2f,    BB  -  L—    =   0. 


.  a;  =  1,    y  —  *• 

8.  2y  +  3«  =  y  +  43,    y  -  ^=—  =  y  -  ^- 

o  o 

Ans.  x  —   10,    y  =  13. 

9.  4y  —  x  ~  y  =  x  +  18,   and   27  —  y  —  X  +  y  +  t. 

2 


.   a;  =  9,    y  =.  7. 

f-2   =|. 

.   a;  =  10,    y  =  20. 

116.  Having  explained  the  principal  methods  of  elimina- 
tion, we  shall  add  a  few  examples  which  may  be  solved  by 
any  one  of  them  ;  and  often  indeed,  it  may  be  advantageous 
to  employ  them  all,  even  in  the  same  example. 

GENERAL      EXAMPLES. 

Find  the  values  of  x  and  y  in  the  following  simultaneous 
equations  : 

1.    2a;  4-  3y  =   16,    and    3x  —  2y   =  11. 

Ans.   x  =  5,    y  =     2. 


ELIMINATION. 


HI 


to       ty  _  j 
2         5          4       -   20' 


3x       2y         61 
T         5"     =  T2~6" 

Ans.  x  =  -,    y  =  - 


3.          +  7y  =  99,    and        +  7s  =  51. 

7  7 


4       *_i2-y 
*       2        "4 


.4ns.   a;  =  7,    y  =   14. 

?L+y  + 1  _  8  =  2_y^  +  27. 


5. 


4x 
5 


x  — 


—  V        x  4-  V 

6.,    -^rV^-8* 


j«- 


7. 


8. 


9. 


3y  -  as       2a;  -  y  _     g 


e  —  y  + 

•  o*K  ~~~    o 

4 
Sx  —  3  — 


o 
6  -  y 


4a;  —  4       y  —  5 


!-i*-iy  + 


=  79 


6  = 


f_ 


Ans.   x  =  CO,    y  •=.  40. 


a;  =  6. 


y-5. 


E  L  E  M  E  N  T  A  li  Y      A  L  G  K  B  K  A  . 

c  +  ab  —  bd 

I  x  = 

Ans. 


IT. 


P  li  O  B  L  K  M  8 .  143 


PROBLEMS. 
1.  What  fraction  is  that,  to  the  numerator  of  which  if  1 

be  added,  the  value  will  be   -  ,  but  if  1   be  added   to  ita 

1 
denominator,  the  value  will  be    -  ? 

m 

Let  the  fraction  be  denoted  by   -  • 

y 

Then,  by  the  conditions, 

x  -f  1         1  x  I 

—  ;»>  :"d'  y  +  i--r 

whence,       Sx  +  3  =  y,     and    4x  =  y  +1. 
Therefore,  by  subtracting, 

x  —  3  =  1,    and    x  =  4. 
Hence,  12  +  3  =  y; 

.-.     y  =  15. 

2.  A  market-woman  bought  a  certain  number  of  eggs  at 
2  for  a  penny,  and  as  many  others  at  3  for  a  penny  ;  and 
having  sold  them  all  together,  at  the  rate  of  5  for  2c?,  found 
that  she  had  lost  4d:  how  many  of  both  kinds  did  she  buy  ? 

Let  2x     denote  the  whole  number  of  eggs. 

Then,  x  —     the  number  of  eggs  of  each  sort. 

Then  will,      -x  =     the  cost  of  the  first  sort, 

•"  •       « 

and,  -x  =     the  cost  of  the  second  sort. 

3 

But,  by  the  conditions  of  the  question, 


hence,    —     will   denote   the   amounts   for  which  the   eggs 

O 

were  sold. 


144  ELEMENTARY      ALGEBBA. 

But,  by  the  conditions, 

1          1    •     4« 

S"+s?:-T--41 

therefore,  15«  -f  10x  —  24ce  =  120; 

.*.     x  =  120  ;  the  number  of  eggs  of  each  sort. 

3.  A  person  possessed  a  capital  of  30,000  dollars,  for 
which  he  received  a  certain  interest  ;  but  he  owed  the  sum 
of  20,000  dollars,  for  which  he  paid  a  certain  annual  interest. 
The  interest  that  he  received  exceeded  that  which  he  paid 
by  800  dollars.  Another  person  possessed  35,000  dollars,  for 
which  he  received  interest  at  the  second  of  the  above  rates  ; 
but  he  owed  24,000  dollars,  for  which  he  paid  interest  at  the 
first  of  the  above  rates.  The  interest  that  he  received,  an- 
nually, exceeded  that  which  he  paid,  by  310  dollars.  Re- 
quired the  two  rates  of  interest. 

Let  x  denote  the  number  of  units  in  the  first  rate  of 
interest,  and  y  the  unit  in  the  second  rate.  Then  each  may 
be  regarded  as  denoting  the  interest  on  $100  for  1  year. 

To  obtain  the  interest  of  $30,000  at  the  first  rate,  denoted 
by  «,  we  form  the  proportion, 

100  :  30,000  :  :  x  :  —  '—-  —  ,  or  300a;. 
And  for  the  interest  of  $20,000,  the  rate  being    y, 


100  :  20,000  :  :  y  :          -        Or  200y. 

But,  by  the  conditions,  the  difference  between  these  two 
amounts  is  equal  to  800  dollars. 

We  have,  then,  for  the  first  equation  of  the  problem, 
SOOaj  —  200y  =   800. 


PROBLEMS.  145 

By  expressing,  algebraically,  the  second  condition  of  the 
problem,  we  obtain  a  second  equation, 

350y  —  240o;  =   310. 

Both  members  of  the  first  equation  being  divisible  by  100, 
and  those  of  the  second  by  10,  we  have, 

$x  —  2y  =   8,         35y  —  24«  =   31. 

To  eliminate  jc,  multiply  the  first  equation  by  8,  and  then 
add  the  result  to  the  second  ;  there  results, 

19y  =  95,     whence,     y  =  5. 

Substituting  for   y,   in  the  first  equation,  this  value,  and 
that  equation  becomes, 

3x  —  10  =  8,     whence,     x  =  6. 
Therefore,  the  first  rate  is  6  per  cent,  and  the  second  5. 

VERIFICATION. 

$30,000,     at  6  per  cent,  gives     30,000  X  .06  =  $1800. 
$20,000,  5         "  "         20,000  X  .05   —  $1000. 

And  we  have,       1800  —  1000  =  800. 

The  second  condition  can  be  verified  in  the  same  manner. 

4.  "What  two  numbers  are  those,  whose  difference  is  7, 
and  sum  33  ?  Ans.  13  and  20. 

5.  Divide  the  number  75  into  two  such  parts,  that  three 
times  the  greater  may  exceed  seven  times  the  less  by  15. 

Ans.  54  and  21. 

6.  In  a  mixture  of  wine  and  cider,  |  of  the  whole  plus  25 
gallons  was  wine,  and  £  part  minus  5  gallons  was  cider :  how 
many  gallons  were  there  of  each  ? 

Ans.  85  of  wine,  and  35  of  cider. 
7 


146  ELEMENTARY      ALGEBRA. 

7.  A  bill  of  £120  was  paid  in  guineas  and  moidores,  and 
the  number  of  pieces  used,  of  both  sorts,  was  just  100.    If 
the  guinea  be  estimated  at  21s,  and  the  moidore  at  2Vs,  how 
many  pieces  were  there  of  each  sort  ?  Ans.  50. 

8.  Two  travelers  set  out  at  the  same  time  from  London 
and  York,  whose  distance  apart  is  150  miles.     One  of  them 
travels  8  miles  a  day,  and  the  other  7 :  in  what  time  will 
they  meet?  Ans.  In  10  days. 

9.  At  a  certain  election,  375  persons  voted  for  two  candi- 
dates, and  the  candidate  chosen  had  a  majority  of  91  :  how 
many  voted  for  oaeli  ? 

Ans.  233  for  one,  and  142  for  the  other. 

10.  A  person  has  two  horses,  and  a  saddle  worth  £50. 
Now,  if  the  saddle  be  put  on  the  back  of  the  first  horse,  it 
makes  their  joint  value  double  that  of  the  second  horse ; 
but  if  it  be  put  on  the  back  of  the  second,  it  makes  their 
joint  value  triple  that  of  the  first :  what  is  the  value  of  each 
horse  ?  Ans.  One  £30,  and  the  other  £40. 

11.  The  hour  and  minute  hands  of  a  clock  are  exactly  to- 
gether at  12  o'clock :  when  will  they  be  again  together? 

Ans.  Ih.  5/Tni' 

12.  A  man  and  his  wife  usually  drank  out  a  cask  of  beer 
in  12  days ;  but  when  the  man  was  from  home,  it  lasted  the 
woman  30  days :  how  many  days  would  the  man  alone  be 
in  drinking  it  ?  Ans.  20  days. 

13.  If  32  pounds  of  sea-water  contain  1  pound  of  salt,  how 
much  fresh  water  must  be  added  to  these  32  pounds,  in  order 
that  the  quantity  of  salt  contained  in  32  pounds  of  the  new 
mixture  shall  be  reduced  to  2  ounces,  or  |  of  a  pound  ? 

Ans.  224  Ibs. 

14.  A  person  who  possessed  100,000  dollars,  placed  the 
greater  part  of  it  out  at  5  per  cent  interest,  and  the  other 


PUOBLKMS.  14:7 

at  4  per  cent.    The  interest  which  he  received  for  the  whole, 
amounted  to  4640  dollars.     Required  the  two  parts. 

Ans.    $64,000  and  $36,000. 

15.  At  the  close  of  an  election,  the  successful  candidate 
had  a  majority  of  1500  votes.  Had  a  fourth  of  the  votes  of 
the  unsuccessful  candidate  been  also  given  to  him,  he  would 
have  received  three  times  as  many  as  his  competitor,  want- 
ing three  thousand  five  hundred  :  how  many  votes  did  each 
receive?  •  i  1st,  6500. 

2d,  5000. 


• 
-    \ 


16.  A  gentleman  bought  a  gold  and  a  silver  watch,  and  a 
chain  worth  $25.  When  he  put  the  chain  on  the  gold  watch» 
it  and  the  chain  became  worth,  three  and  a  half  times  more 
than  the  silver  watch  ;  but  when  he  put  the  chain  on  the 
silver  watch,  they  became  worth  one-half  the  gold  watch 
and  15  dollars  over  :  what  was  the  value  of  each  watch  ? 

j  Gold  watch,  $80. 
(Silver     «       $30. 

1  7.  There  is  a  certain  number  expressed  by  two  figures, 
which  figures  are  called  digits.  The  sum  of  the  digits  is  11, 
and  if  13  be  added  to  the  first  digit  the  sum  will  be  three 
times  the  second:  what  is  the  number?  Ans.  56. 

18.  From  a  company  of  ladies  and  gentlemen  15  ladies 
retire;    there  are  then  left  two  gentlemen  to  each  lady. 
After  which  45  gentlemen  depart,  when  there  are  left  5 
ladies  to  each  gentleman  :  how  many  were  there  of  each  at 

first  ?  (  50  gentlemen. 

Ans.   1  .    f  ,. 

}  40  ladies. 

19.  A  person  wishes  to  dispose  of  his  horse  by  lottery. 
If  he  sells  the  tickets  at  $2  each,  he  will  lose  $30  on  his 
horse  ;  but  if  he  sells  them  at  $3  each,  he  will  receive  $30 


14:8  KLKMKNTAIIY       ALGEBRA. 

more  than  his  horse  cost  him.    "What  is  the  value  of  the 
horse,  and  number  of  tickets?        .         j  Horse,  $150. 

'    (No.  of  tickets,  CO. 

20.  A  person  purchases  a  lot  of  wheat  at  $1,  and  a  lot  of 
rye  at  75  cents  per  bushel ;  the  whole  costing  him  §1 17.50. 
He  then  sells  £  of  his  wheat  and  1  of  his  rye  at  the  same  rate, 
and  realizes  $27.50.  How  much  did  he  buy  of  each  ? 

80  bush,  of  wheat. 


Ans.    -,  ,       , 

50  bush,  of  rye. 

21.  There  are  52  pieces  of  money  in  each  of  two  bags.     A. 
takes  from  one,  and  IB  from  the  other.     A  takes  twice  as 
much  as  J3  left,  and  J?  takes  7  times  as  much  as  A  left. 

How  much  did  each  take?  (  A.  48  pieces. 

Ans.   <      ' 

(  B,  28  pieces. 

22.  Two  persons,  A  and  B,  purchase  a  house  together, 
Worth  $1200.     Says  A  to  _Z?,  give  me  two-thirds  of  your 
money  and  I  can  purchase  it  alone ;  but,  says  J5  to  A,  if 
you  will  give  me  three-fourths  of  your  money  I  shall  be  able 
to  purchase  it  alone.     How  much  had  each  ? 

Ans.   A,  $800 ;  J5,  $600. 

23.  A  grocer  finds  that  if  he  mixes  sherry  and  brandy  in 
the  proportion  of  2  to  1,  the  mixture  will  be  worth  78s.  per 
dozen ;  but  if  he  mixes  them  in  the  proportion  of  7  to  2,  he 
can  get  79s.  a  dozen.     "What  is  the  price  of  each  liquor  per 
dozen?  Ans.   Sherry,  81s. ;  brandy,  72s. 


Equations  containing  three  or  more  unknown  quantities. 


Let  us  now  consider  equations  involving  three  or 
more  unknown  quantities. 

Take  the  group  of  simultaneous  equations, 

117.  Give  the  rule  for  solving  any  group  of  simultaneous  equations? 


EXAMPLES.  149 

5x  -  Qy  +  42  =   15,  .     .       (1.) 

7a  +  4y  —  3z  =   19,  .     .       (2.) 

2x  +    y  -f-  62  =  46.      ...       (3.) 

To  eliminate  z  by  means  of  the  first  two  equations,  multi- 
ply the  first  by  3,  and  the  second  by  4 ;  then,  since  the 
coefficients  of  z  have  contrary  signs,  add  the  two  results 
together.  This  gives  a  new  equation : 

43z  —  2y  =   121  .     .     .     .     .     (4.) 

Multiplying  the  second  equation  by  2  (a  factor  of  the 
coefficient  of  z  in  the  third  equation),  and  adding  the  result 
to  the  third  equation,  we  have, 

16a  +  9y  =  84 '   (5.) 

The  question  is  then  reduced  to  finding  the  values  of  x 
and  y,  which  will  satisfy  the  new  Equations  (4)  and  (5). 

Now,  if  the  first  be  multiplied  by  9,  the  second  by  2,  and 
the  results  added  together,  we  find, 

419a;  =   1257;  whence,    x  =  3. 

"We  might,  by  means  of  Equations  ( 4 )  and  ( 5 )  deter- 
mine y  in  the  same  way  that  we  have  determined  x ;  but 
the  value  of  y  may  be  determined  more  simply,  by  substi- 
tuting the  value  of  x  in  Equation  ( 5  ) ;  thus, 

04     40 

48  +  Qy  =   84.  .-.     y  =  -  — =  4. 

In  the  same  manner,  the  first  of  the  three  given  equations 
becomes,  by  substituting  the  values  of  x  and  y, 

15  —  24  -f  4s  =  15.  .'.    z  =  —  =  6. 

4 

In  the  same  way,  any  group  of  simultaneous  equations 
may  be  solved.  Hence,  the 


150  ELEMENTARY      ALGEBRA. 

EULE. 

1.  Combine  one  equation  of  the  group  icith  each  of  the 
others,  by  eliminating  one  unknown  quantity;  there  will 
result  a  neio  group  containing  one  equation  less  than  the 
original  group  : 

II.  Combine  one  equation  of  this  new  group  with  each 
of  the  others,  by  eliminating  a  second  unknown  quantity  ; 
there  will  result  a  new  group  containing  two  equations  less 
than  the  original  group : 

HI.  Continue  the  operation  until  a  single  equation  is 
found,  containing  but  one  unknotcn  quantity : 

IV.  find  the  value  of  this  unknown,  quantity  by  the 
preceding  rules  ;  substitute  this  in  one  of  the  group  of 
two  equations,  and  find  tlie  value  of  a  second  unknown 
quantity  ;  substitute  these  in  either  of  the  group  of  three, 
finding  a  third  unknown  quantity  ;  and  so  on,  till  the 
values  of  all  are  found. 

NOTES. — 1.  lu  order  that  the  value  of  the  unknown  quan- 
tities may  be  determined,  there  must  be  just  as  many  inde- 
pendent equations  of  condition  as  there  are  unknown  quan- 
tities. If  there  are  fewer  equations  than  unknown  quantities, 
the  resulting  equation  will  contain  at  least  two  unknown 
quantities,  and  hence,  their  values  cannot  be  found  (Art.  110). 
If  there  are  more  equations  than  unknown  quantities,  the 
conditions  maybe  contradictory,  and  the  equations  impossible. 

2.  It  often  happens  that  each  of  the  proposed  equations 
does  not  contain  all  the  unknown  quantities.     In  this  case, 
with  a  little  address,  the  elimination  is  very  quickly  per- 
formed. 

Take  the  four  equations  involving  four  unknown  quanti- 
ties : 

2x  —  3y  +  2z  =   13.     (1.)  4y  +  2z  =   14.     (3.) 

4u  —  2x  =  30.     (2.)  5y  +  3u  —   32.     (4.) 


EXAMPLES.  151 

By  inspecting  these  equations,  we  see  that  the  elimination 
of  s  in  the  two  Equations,  ( 1 )  and  (  3 ),  will  give  an  equa- 
tion involving  x  and  y\  and  if  we  eliminate  u  in  Equa- 
tions ( 2 )  and  (  4 ),  we  shall  obtain  a  second  equation,  in- 
volving x  and  y.  These  last  two  unknown  quantities  may 
therefore  be  easily  determined.  In  the  first  place,  the 
elimination  of  z  from  ( 1 )  and  ( 3 )  gives, 

ty  -  2x  =  1 ; 

That  of  u  from  ( 2 )  and  ( 4  )  gives, 
20y  +  Gx  =  38. 
Multiplying  the  first  of  these  equations  by  3,  and  adding, 

41y  =  41; 

Whence,  y  =     1. 

Substituting  this  value  in    1y  —  2x  =  1,    we  find, 

x  =  3. 

Substituting  for  x  its  value  in  Equation  ( 2 ),  it  becomes 

4u  —  6  =  30. 

Whence,  u  =  9. 

And  substituting  for   y   its  value  in  Equation  (8),  there 

results, 

z  —  5. 

EXAMPLES. 

X  +      y  +      2    =    29 

x  +  2y  +  3z  =  62 
1.  Given  <  >  to  find  «,  y,  and  z. 

+  ^  +  f  =  :o 

Ans.  x  =  8,    y  =  9,    z  —  12. 


152 


ELEMENT AKY       ALGEBRA. 


f  2x  +  4y  —  33  =   22  ~\ 

2.  Given  <  4x  —  2y  +  5z  —  18   I  to  find  cr,  y,  and  a. 
[  6z  +  7y  -    z  =  63  J 

Ans.  x  =  3,    y  =  7,    z  =  4. 


3.  Given  • 


^B  +  -V  4-  -8  = 


L  f 


to  find  a?,  y,  and  z. 


2  =  12 

.   a;  =   12,    y  =  20,    z  =  30. 


4.  Given  <  x  +  y  —  z  =  18$   >  to  find  a,  y,  and  z. 
x  -  y  +  z  =  13|J 

^ln«.   x  =  16,    y  —   V£,    s  =  5^ 

{3a;  +  5y  =   161  ^ 
7*  +  23    =  209   V  to  find  a,  y,  and  2. 
2y  +    z    -     89  J 

.   x  =  17,    y  =  22,    z  =  45. 


6.  Given  •« 


x  = 


1        1 

— I —   =  a 
x       y 

-1  +  1  =  6 

X          Z 


to  find  a;,  y,  and  z. 


a  +  b  —  c' 


a  +  c  —  £' 


z  = 


b+c  — 


NOTE. — In  this  example  we  should  not  proceed  to  clear 
the  equation  of  fractions;  but  subtract  immediately  the 
second  equation  from  the  first,  and  then  add  the  third :  we 
thus  find  the  value  of  y. 


PROBLEMS.  153 


PKOBLEMS. 

1.  Divide  the  number  90  into  four  such  parts,  that  the 
first  increased  by  2,  the  second  diminished  by  2,  the  third 
multiplied  by  2,  and  the  fourth  divided  by  2,  shall  be  equal 
each  to  each. 

This  problem  may  be  easily  solved  by  introducing  a  new 
unknown  quantity. 

Let  a;,  y,  z,  and  u,  denote  the  required  parts,  and  desig- 
nate by  m  the  several  equal  quantities  which  arise  from  the 
conditions.  We  shall  then  have, 

u 

x  +  2  =  m,     y  —  2  =  m,     2«  =  m,     -   =  m. 

2 

From  which  we  find, 

x  =  m  —  2,     y  =  m  +  2,     z  =  — ,     u  =  2m. 
And,  by  adding  the  equations, 

x  -{-  y  -\-  z  +  u  =  m  -f  m  H -f  2m  =  4£m. 

And  since,  by  the  conditions  of  the  problem,  the  first 
member  is  equal  to  90,  we  have, 

4|m  =   90,     or     fw  —  90; 
hence,  m  =  20. 

Having  the  value  of  m,  we  easily  find  the  other  values; 

viz.: 

a;  —   18,    y  =  22,     z  =  10,     u  =  40. 

2.  There  are  three  ingots,  composed  of  different  metals 
mixed  together.    A  pound  of  the  first  contains  7  ounces  of 
silver,  3  ounces  of  copper,  and  6  of  pewter.     A  pound  of 
the  second  contains  12  ounces  of  silver,  3  ounces  of  copper, 
and  1  of  pewter.     A  pound  of  the  third  contains  4  ounces 
of  silver,  7  ounces  of  copper,  and  5  of  pewter.     It  is  required 

7* 


EL  K  M  K  x  T  A  11  y     A  L  G  K  B  K  A . 

to  find  how  much  it  will  take  of  each  of  the  three  ingots  to 
form  a  fourth,  which  shall  contain  in  a  pound,  8  ounces  of 
silver,  3J  of  copper,  and  4£  of  pewter. 

Let  a,  y,  and  2,  denote  the  number  of  ounces  which  it 
is  necessary  to  take  from  the  three  ingots  respectively,  in 
order  to  form  a  pound  of  the  required  ingot.  Since  there 
are  7  ounces  of  silver  in  a  pound,  or  16  ounces,  of  the  first 
ingot,  it  follows  that  one  ounce  of  it  contains  T^-  of  an  ounce 
of  silver,  and,  consequently,  in  a  number  of  ounces  denoted 

1x 

by  a*,  there  is  --  ounces  of  silver.     In  the  same  manner, 
16 

12v  4  z 

we  find  that,  -  — ,  and  — ,  denote  the  number  of  ounces 
16  16 

of  silver  taken  from  the  second  and  third ;  but,  from  the 
enunciation,  one  pound  of  the  fourth  ingot  contains  8  ounces 
of  silver.  We  have,  then,  for  the  first  equation, 

"tx       12  y       42 
16  4'  T6~  "  16   = 

or,  clearing  fractions, 

Va;  +  12y  +  42  =   128. 
As  respects  the  copper,  we  should  find, 
3aj  -f  3y  +  72  =  60 ; 
and  with  reference  to  the  pewter, 

6«  +  y  +  02  =  68. 

As  the  coefficients  of  y  in  these  three  equations  are  the 
most  simple,  it  is  convenient  to  eliminate  this  unknown 
quantity  first. 

Multiplying  the  second  equation  by  4,  and  subtracting  the 
first  from  it,  member  from  member,  we  have, 

5x  +  242  r=   112. 


PROBLEMS.  15ft 

Multiplying  the  third  equation  by  3,  and  subtracting  the 
second  from  the  resulting  equation,  we  have, 

I5x  +  Sz  =   144. 

Multiplying  this  last  equation  by  3,  and  subtracting  the 
preceding  one,  we  obtain, 

40a;  =  320; 
whence,  x  =  8. 

Substitute  this  value  for  x  in  the  equation, 

I5x  +  Sz  =   144; 

it  becomes,  120  -f  8z  =  144, 

whence,  2  =  3. 

Lastly,  the  two  values,  x  =  8,  z  =  3,  being  substituted 
in  the  equation, 

6aj  +  y  +  5z  =  68, 
give,  48  +  y  +  15  =  68, 

whence,  y  =     5. 

Therefore,  in  order  to  form  a  pound  of  the  fourth  ingot, 
we  must  take  8  ounces  of  the  first,  5  ounces  of  the  second, 
and  3  of  the  third. 

VERIFICATION. 

If  there  be  1  ounces  of  silver  in  16  ounces  of  the  first 
ingot,  in  eight  ounces  of  it  there  should  be  a  number  of 
ounces  of  silver  expressed  by 

7x8 

16 
In  like  manner, 

12  x  5  ,4x3 

onri       _ 

~T6      '    '  16     ' 

will  express  the  quantity  of  silver  contained  in  5  ounces  of 
the  second  ingot,  and  3  ounces  of  the  third. 


156  E  L  K  M  E  N  T  A  K  Y       ALGEBRA. 

Now,  we  have, 

7X8        12   X  5        4x3          128 
16  16  16  16 

therefore,  a  pound  of  the  fourth  ingot  contains  8  oiui  ;es  of 
silver,  as  required  by  the  enunciation.  The  same  conditions 
may  be  verified  with  respect  to  the  copper  and  pewter. 

3.  A>s  age  is  double  .Z?'s,  and  IPs  is  triple  of  <7's,  and  the 
sum  of  all  their  ages  is  140 :  what  is  the  age  of  each? 

Ans.  A's  -  84;  B's  =  42;  and  C's  =  14. 

4.  A  person  bought  a  chaise,  horse,  and  harness,  for  £60 ; 
the  horse  came  to  twice  the  price  of  the  harness,  and  the 
chaise  to  twice  the  cost  of  the  horse  and  harness  :  what  did 
he  give  for  each?  (  £13    6s.  8d.  for  the  horse. 

Ans.  i    £6  135.  4d.  for  the  harness. 
(  £40  for  the  chaise. 

5.  Divide  the  number  36  into  three  such  parts  that  1  of 
the  first,  i  of  the  second,  and  1  of  the  third,  may  be  all 
equal  to  each  other.  Ans.  8,  12,  and  16. 

6.  If  A  and  B  together  can  do  a  piece  of  work  in  8  days, 
A  and  C  together  in  9  days,  and  B  and  C  in  ten  days,  how 
many  days  would  it  take  each  to  perform  the  same  work 
alone?  Ans.   A,  14|f ;  B,  I7ff;   C,  23/T. 

7.  Three  persons.  A,  B,  and  (7,  begin  to  play  together, 
having  among  them  all  $600.     At  the  end  of  the  first  game 
A  has  won  one-half  of  _B's  money,  which,  added  to  his  own, 
makes  double  the  amount  B  had  at  first.     In  the  second 
game,  A  loses  and  B  wins  just  as  much  as  C  had  at  the  be- 
ginning, when  A  leaves  off  with  exactly  what  he  had  at  first : 
how  much  had  each  at  the  beginning  ? 

Ans.    A,  $300  ;  B,  §200  ;   C  $100. 

8.  Three  persons,  A,  B,  and  C,  together  possess  $3640. 


PROBLEMS.  157 

If  B  gives  A  $400  of  his  money,  then  A  will  have  $320 
more  than  B\  but  if  B  takes  §140  of  C 's  money,  then  B 
and  C  will  have  equal  sums  :  how  much  has  each  ? 

Ans.   A,  $800  ;  B,  $1280;   <7,  $1560. 

9.  Three  persons  have  a  bill  to  pay,  which  neither  alone 
is  able  to  discharge.     A  says  to  B,  "  Give  me  the,  4th  of 
your  money,  and  then  I  can  pay  the  bill."     B  says  to   (7, 
"  Give  me  the  8th  of  yours,  and  I  can  pay  it."     But  C  says 
to  A)  "  You  must  give  me  the  half  of  yours  before  I  can 
pay  it,  as  I  have  but  $8  "  :  what  was  the  amount  of  their 
bill,  and  how  much  money  had  A  and  B  ? 

.        (  Amount  of  the  bill,  $13. 
S'  (A  had  $10,    and  .B  $12. 

10.  A  person  possessed  a  certain  capital,  which  he  placed 
out  at  a  certain  interest.     Another  person,  who  possessed 
10000  dollars  more  than  the  first,  and  who  put  out  his  capital 
1  per  cent,  more  advantageously,  had  an  annual   income 
greater  by  800   dollars.     A  third  person,  who  possessed 
15000  dollars  more  than  the  first,  putting  out  his  capital  2 
per  cent,  more  advantageously,  had  an  annual  income  greater 
by  1500  dollars.     Required,  the  capitals  of  the  three  per- 
sons, and  the  rates  of  interest. 

.       j  Sums  at  interest,   $30000,  $40000,  $45000. 
'  (  Rates  of  interest,        4  5  6  pr.  ct. 

11.  A  widow  receives  an  estate  of  $15000  from  her  de- 
ceased husband,  with  directions  to  divide  it  among  two  sons 
and  three  daughters,  so  that  each  son  may  receive  twice  as 
much  as  each  daughter,  and  she  herself  to  receive  $1000 
more  than  all  the  children  together :  what  was  her  share, 
and  what  the  share  of  each  child  ? 

(  The  widow's  share,  $8000 

Ans.  <  Each  son's,  $2000 

'  Each  daughter's,      $1000 


158  ELEMENTARY       ALGKBRA. 

12.  A  certain  sum  of  money  is  to  be  divided  between 
three  persons,  A,  B,  and   C.    A  is  to  receive  $3000  les? 
than  half  of  it,  B  $1000  less  than  one-third  part,  and   C  to 
receive  $800  more  than  the  fourth  part  of  the  whole  :  what 
is  the  sum  to  be  divided,  and  what  does  each  receive  ? 

{Sum,  $38400. 

A  receives  $16200. 
B  "  $11800. 
G  "  $10400. 

13.  A  person  has  three  horses,  and  a  saddle  which  is  worth 
$220.     If  the  saddle  be  put  on  the  back  of  the  first  horse,  it 
will  make  his  value  equal  to  that  of  the  second  and  third  ; 
if  it  be  put  on  the  back  of  the  second,  it  will  make  his  value 
double  that  of  the  first  and  third  ;  if  it  be  put  on  the  back 
of  the  third,  it  will  make  his  value  triple  that  of  the  first 
and  second :  what  is  the  value  of  each  horse  ? 

Ans.    1st,  $20 ;    2d,  $100  ;    3d,  $140. 

14.  The  crew  of  a  ship  consisted  of  her  complement  of 
sailors,  and  a  number  of  soldiers.    There  wrere  22  sailors  to 
every  three  guns,  and  10  over ;  also,  the  whole  number  of 
hands  was  five  times  the  number  of  soldiers  and  guns  to- 
gether.   But  after  an  engagement,  in  which  the  slain  were 
one-fourth  of  the  survivors,  there  wanted  5  men  to  make 
13  men  to  every  two  guns:  required,  the  number  of  guns, 
soldiers  and  sailors. 

Ans.  90  guns,  55  soldiers,  and  670  sailors. 

15.  Three  persons  have  $96,  which  they  wish  to  divide 
equally  between  them.     In  order  to  do  this,  A,  who  has  the 
most,  gives  to  B  and  C  as  much  as  they  have  already ;  then 
B  divides  with  A  and  C  in  the  same  manner,  that  is,  by 
giving  to  each  as  much  as  he  had  after  A  had  divided  with 
them  •   C  then  makes  a  division  with  A  and  J9,  when  it  is 


PROBLK  MS.  159 

found  that  they  all  have  equal  sums:  how  ui.ich  had  each 
at  first?  Ans.  1st,  $52;    2d,  $28;    3d,  $16. 

16.  Divide  the  number  a  into  three  such  parts,  that  the 
first  shall  be  to  the  second  as  m  to  n,  and  the  second  to  the 
third  as  p  to  q. 

amp  anp  anq 


mp+np-rnq  mp+np+nq  mp+np+nq 

17.  Three  masons,  A,  J3,  and  C",  are  to  build  a  wall.  A 
and  B  together  can  do  it  in  12  days ;  B  and  C  in  20  days  ; 
and  A  and  C  in  15  days :  in  what  time  can  each  do  it  alone, 
and  in  what  time  can  they  all  do  it  if  they  work  together  ? 

Ans.   A,  in  20  days ;   J5,  in  30  ;   and  C,  in  60 ;  all,  in  10. 


160  ELEMENTARY      ALGEBRA. 


CHAPTER  VI. 

FORMATION   OF  POWERS. 

118.  A  POWER  of  a  quantity  is  the  product  obtained  by 
taking  that  quantity  any  number  of  times  as  a  factor. 

If  the  quantity  be  taken  once  as  a  factor,  we  have  the  first 
power ;  if  taken  twice,  we  have  the  second  power ;  if  three 
tunes,  the  third  power;  if  n  tunes,  the  nth  power,  n  being 
any  whole  number  whatever. 

A  power  is  indicated  by  means  of  the  exponential  sign 

thus, 

a  =  a1   denotes  first  power  of  a.* 

axa  =  a?  "  square,  or  2d  power  of  a. 

a  X  a  x  a  =  a3  "  cube,  or  third  power  of  a. 

axaxaxa  =  a*  "  fourth  power  of  a. 

axaxaxaxa  =  a5  "  fifth  power  of  a. 

axaxaxa —  =  am  "  m'*  power  of  a. 

In  every  power  there  are  three  things  to  be  considered : 

1st.  The  quantity  which  enters  as  a  factor,  and  which  is 
called  the  first  power. 

2d.  The  small  figure  which  is  placed  at  the  right,  and 
a  little  above  the  letter,  is  called  the  exponent  of  the 

» Since  a°  =  1  (Art.  49),  a9  X  a  =  1  X  a  =  a1 ;  so  that  the  two 
factors  of  a1,  are  1  and  a. 

118.  What  is  a  power  of  a  quantity?  What  is  the  power  when  the 
quantity  is  taken  once  as  a  factor  ?  When  taken  twice  ?  Three  times  ? 
n  times  ?  How  is  a  power  indicated  ?  In  every  power,  how  many  things 
are  considered  ?  Name  them. 


POWERS      OF      MONOMIALS.  161 

power,  and  shows  how  many  times  the  letter  enters  as  a 
factor. 

3d.  The  power  itself,  which  is  the  final  product,  or  result 
of  the  multiplications. 

POWERS      OF     MONOMIALS. 

1  19.     Let  it  be  required  to  raise  the  monomial   2a3£>2  to 
the  fourth  power.     We  have, 

(2a3S2)4  =  2a352  x  2a3bz  X  2a?I>*  X  2a3&2, 

which  merely  expresses  that  the  fourth  power  is  equal  to 
the  product  which  arises  from  taking  the  quantity  four 
times  as  a  factor.  By  the  rules  for  multiplication,  this  pro- 
duct is 


from  which  we  see, 

1st.  That  the  coefficient  2  must  be  raised  to  the  4th 
power  ;  and, 

2d.  That  the  exponent  of  each  letter  must  be  multiplied 
by  4,  the  exponent  of  the  power. 

As  the  same  reasoning  applies  to  every  example,  we  have, 
for  the  raising  of  monomials  to  any  power,  the  following 

RULE. 

I.  JRaise  the  coefficient  to  the  required  power  : 
IT.  Multiply  the  exponent  of  each  letter  by  the  exponent 
of  the  power. 

EXAMPLES. 

1.  What  is  the  square  of  3o2y3?  Ans.   Oa4^6. 

119.  What  is  the  rule  for  raising  a  monomial  to  any  power?  When 
the  monomial  is  positive,  what  will  be  the  sign  of  its  powers  ?  When 
negative,  what  powers  will  be  plus?  what  minus? 


162  ELEMENTARY      ALGEBRA. 

2.  What  is  the  cube  of  6a5y-x  ?  Ans.   216aiyx3. 

3.  What  is  the  fourth  power  of  2aV** ?          16a12y12£20. 

4.  What  is  the  square  of  a?bsy3  ?  ^4.ns.  a4510y6. 

5.  What  is  the  seventh  power  of  a?bcd3? 

Ans. 

6.  What  is  the  sixth  power  of  a2£3c2<?? 

Ans. 

7.  What  is  the  square  and  cube  of    —  2a2J2  ? 

Square.  Cube. 

—  2azb2 


By  observing  the  way  in  which  the  powers  are  formed, 
we  may  conclude, 

1st.  When  the  monomial  is  positive^  all  the,  powers  icill 
be  positive. 

2d.  When  the  monomial  is  negative,  all  even  powers  icill 
be  positive,  and  all  odd  will  be  negative. 

8.  What  is  the  square  of  —  2a*b5  ?  Ans.   4a8£10. 

9.  What  is  the  cube  of  —  5anb2  ?        Ans.    —  125a3nb*. 

TO.  What  is  the  eighth  power  of  —  «3a;y2  ? 

Ans.    +'a2'ix8y16. 

11.  What  is  the  seventh  power  of  —  ambnc  ? 

Ans.    —  a!mb~*cl. 

12.  Whot  is  the  sixth  power  of  2ab6y5  ? 

Ans. 


P  O  W  E  B  8      OF      F  li  A  C  T  I  O  N  S  .  163 

13.  What  is  the  ninth  power  of  —  anbci  ? 

Ans.    —  a9n69c18. 

14.  What  is  the  sixth  power  of  —  Sabzd? 

Ans.   729a6J12d8. 

15.  What  is  the  square  of  —  10amZ»nc3  ? 

Ans.    100a2m62nc6. 

16.  What  is  the  crie  of  —  9ambnd3f*  ? 

Ans.    — 

17.  What  is  the  fourth  power  of  —  4aflft 

Ans.   256a20£12c16c?20. 

18.  What  is  the  cube  of  —  4«2m52nc3dr? 

Ans.    — 

19.  What  is  the  fifth  power  oflaWxy  ? 

Ans. 

20.  What  is  the  square  of  20xni/mc5?    Ans.   400a52ny2mc10. 

21.  What  is  the  fourth  power  of  3an£2"c3? 

Ans. 

22.  What  is  the  fifth  power  of  —  cndZmy?\f  ? 

Ans.    —  c5n 

23.  What  is  the  sixth  power  of  —  c^b2"?*  ? 

Ans. 

24.  What  is  the  fourth  power  of  —  2azc2d3. 

Ans. 


POWERS       OF      FRACTIONS. 

1S4O.     From  the  definition  of  a  power,  and  the  rule  for 
the  multiplication  of  fractions,  the  cube  of  the  fraction  ^,  is 

written, 

(a\3  _  a       a       a  _    a? 
b)  ~-~~  b  x  b  x  b  ==  ^  ; 

120.  What  is  the  rule  for  raising  a  fraction  to  any  power* 


16:1  ELEMENTARY       ALGEBRA. 

and  since  any  fraction  raised  to  any  power,  may  be  written 
under  the  same  form,  we  find  any  power  of  a  fraction  by 
the  following 

BT7LE. 

Raise  the  numerator  to  the  required  power  for  a  new 
numerator,  and  the  denominator  to  the  required  power  for 
a  new  denominator, 

The  rule  for  signs  is  the  same  as  in  the  last  article. 
EXAMPLES 

Find  the  powers  of  the  following  fractions : 

a  —  c  \z  a2  —  2ac  +  c2 


/a-_cV 

\b+c) 


1.      ^— -    •  AM. 


b2  +  2bc  +  c2 


2      lXy\. 
2'    I 


,3Jc/ 


3.    1-gM-  ^«. 


4.    (-fp-j-  -4w-   ^ 


M 


cfa\3 


5.    (  — r-^1-  ^s.    — 

6-  (1 

/      3aw4  . 

1  —  ^vf-  I  •  -4ns. 


8.    Fourth  power  of   — — -  •  Ans. 


9.    Cube  of  X- y-- 

x  H-  y  ce3  +  oxly  -{- 


POWERS      OF       BINOMIALS.  165 


2  (Zmx^  ct^  mx^  n 

10.    Fourth  power  of          -•  Ans.    -  -—  - 

** 


„.  f  , 
11.    Fifth  power  of    —  -      —  •  Ans.    — 


POWERS       OF      BINOMIALS. 

121.    A  Binomial,  like  a  monomial,  may  be  raised  to  any 
power  by  the  process  of  continued  multiplication. 

1.  Find  the  fifth  power  of  the  binomial    a  4-  ft. 
a  4-    ft      ...»  .......     1st  power. 

a  4-    b 
a2  4-    ab 

4-    ab  +  ft2 


a  4-    ft 

a3  4-  2a2ft  4- 

aft2 
2«ft2   4-    ft3 

a3  4-  3a2ft  4- 
a  4-    ft 

3aft2   +    ft3    .     . 

.     .     3d  power. 

a*  4~  3a3ft  4~ 

3a2ft2  4-    aft3 
3a2ft2  4-    3aft3   4- 

»     "  ' 

a*  4-  4a3ft  4- 
a  4-    ft 

6a2ft2  4~    4aft3   4" 

ft4       4th  power. 

a5  +  4a4ft  +    6a3ft2  +    4a2ft3  +    aft* 

+    a4ft  +    4a3ft2  +    6a2ft3  +  4aft*  +  ft8 
a8  +  5a4ft  4-  10a3ft2  4-  10a2ft3  4-  5aft*  4-  ft5 


121.  How  may  a  binomial  be  raised  to  any  power? 

122.  How  does  the  number  of  multiplications  compare  with  the  ex- 
ponent of  the  power?     If  the  exponent  is  4,  what  is  the  number  of 
multiplications?     How  many  when  it  is  ?n?     How  many  things  are  con« 
•idorod  in  the  raising  of  powers  ?     Xame  them. 


166  ELEMENTARY       ALOE  BE  A. 

NOTE. — 122.  It  Avill  be  observed  that  tbe  number  of 
multiplications  is  always  1  less  than  the  units  in  the  expo- 
nent of  the  power.  Thus,  if  the  exponent  is  1,  no  multipli- 
cation is  necessary.  If  it  is  2,  we  multiply  once  ;  if  it  is  3, 
twice  ;  if  4,  three  times,  &c.  The  powers  of  polynomials 
may  be  expressed  by  means  of  an  exponent.  Thus,  to 
express  that  a  +  b  is  to  be  raised  to  the  5th  power,  we 
write 

(<•  +  *)5; 

if  to  the  mih  power,  we  write 

(a  +  b}m. 
2.  Find  the  5th  power  of  the  binomial  a  —  b. 

a   —    b 1st  power. 

a  -    b 


a?  —    ab 

—    ab   +  b2 

a?  —  lab   +  b2 2d  power. 

a  —      b 


1^   -       2  ($"J)     \r       Cllfi 

—    a*b  +    2ab*   —  b3 


_  53     ....     3d  power. 
a  -  b  __ 
a*  —  3a3b  +    3azb2  —      ab3 


4a53  +  b*    .    4th  power. 
a  -  b 


-f  4a5*  -  b* 


10a362  —  10a2i3  +  Sab*  —  b5        Ans. 


POWEH8      OF      BINOMIALS.  167 

In  the  same  way  the  higher  powers  may  be  obtained.  By 
examining  the  powers  of  these  binomials,  it  is  plain  that  four 
things  must  be  considered : 

1st.  The  number  of  terms  of  the  power. 
2d.    The  signs  of  the  terms. 
3d.   The  exponents  of  the  letters. 
4th.  The  coefficients  of  the  terms. 

Let  us  see  according  to  what  laws  these  are  formed. 


Of  the  Terms. 

123.  By  examining  the  several  multiplications,  we  shall 
observe  that  the  first  power  of  a  binomial  contains  two  terms ; 
the  second  power,  three  terms ;  the  third  power,  four  terms ; 
the  fourth  power,  five ;  the  fifth  power,  six,  &c. ;  and  hence 
we  may  conclude : 

That  the  number  of  terms  in  any  power  of  a  binomial, 
is  greater  by  one  than  the  exponent  of  the  power. 

Of  the  Signs  of  tJie  Terms. 

124.  It  is  evident  that  when  both  terms  of  the  given 
binomial  are  plus,  all  the  terms  of  the  power  icill  be  plus. 

If  the  second  term  of  the  binomial  is  negative,  then  all 
the  odd  terms,  counted  from  the  left,  will  be  positive,  and 
all  the  even  terms  negative. 


123.  How  many  terms  does  the  first  power  of  a  binomial  contain?    The 
second?     The  third?     The  nth  power? 

•  124.  If  both  terms  of  a  binomial  are  positive,  what  will  be  the  signs 
of  the  terms  of  the  power?  If  the  second  term  is  negative,  how  are  the 
signs  of  the  terms  ? 


168  ELEMENTARY      ALGEBRA. 


Of  the  Exponents. 

125.  The  letter  which  occupies  the  first  place  in  a  bino- 
mial, is  called  the  leading  letter.  Thus,  a  is  the  leading 
letter  in  the  binomials  a  +  ft,  and  a  —  b. 

1st.  It  is  evident  that  the  exponent  of  the  leading  letter 
in  the  first  term,  will  be  the  same  as  the  exponent  of  the 
power;  and  that  this  exponent  will  diminish  by  one  in  each 
term  to  the  right,  until  we  reach  the  last  term,  when  it  will 
be  0  (Art.  49). 

2d.  The  exponent  of  the  second  letter  is  0  in  the  first 
terra,  and  increases  by  one  in  each  term  to  the  right,  to  the 
last  term,  when  the  exponent  is  the  same  as  that  of  the  given 
power. 

3d.  The  sum  of  the  exponents  of  the  two  letters,  in  any 
term,  is  equal  to  the  exponent  of  the  given  power.  This 
last  remark  will  enable  us  to  verify  any  result  obtained  by 
means  of  the  binomial  formula. 

Let  us  now  apply  these  principles  in  the  two  following 
examples,  in  which  the  coefficients  are  omitted : 

(a  +  b)6  .   .   .  a6  +  «5ft  +  «4ft2  +  «3ft3  +  «2ft4  -f  aft5  +  b6, 
(a  —  by  .   .  .a6  -  a5b  +  a*ft2  —  a3ft3  +  a?b*  —  aft5  +  ft6. 

As  the  pupil  should  be  practised  in  writing  the  terms  with 
their  proper  signs,  without  the  coefficients,  we  will  add  a 
few  more  examples. 

125.  Which  is  the  leading  letter  of  a  binomial?  What  is  the  exponent 
of  this  letter  in  the  first  term  ?  How  does  it  change  in  the  terms  towards 
the  right  ?  What  is  the  exponent  of  the  second  letter  in  the  second  term  ? 
How  does  it  change  iu  the  terms  towards  the  right  ?  What  is  it  in  th« 
last  term  ?  What  is  the  sum  of  the  exoonents  in  any  term  equal  to  ? 


POWERS      OF      BINOMIALS.  169 

1.  (a-f  b)3 .  .  «3-fa26+a&2.f    b\ 

2.  (a  —  by  .  .  a*-a3b+a?bz—  ab3  +  b*. 

3.  (a  +  b}5  •  •  as+a*&+a3&2+a2&3+aft*  +    b5. 

4.  (a  -  b)1  .  .  a?-aebi-a5b'i—atb3+a:ibl—azb5+ab*—b'1. 

Of  the  Coefficients. 

126.  The  coefficient  of  the  first  terra  is  1.  The  coeffi 
cient  of  the  second  term  is  the  same  as  the  exponent  of  the 
given  power.  The  coefficient  of  the  third  term  is  found  by 
multiplying  the  coefficient  of  the  second  term  by  the  expo- 
nent of  the  leading  letter  in  that  term,  and  dividing  the 
product  by  2.  And  finally : 

If  the  coefficient  of  any  term  be  multiplied  by  the  expo- 
nent of  the  leading  letter  in  that  term,  and  the  product 
divided  by  the  number  which  marks  the  place  of  the  term 
from  the  left,  the  quotient  will  be  the  coefficient  of  t/te 
next  term. 

Thus,  to  find  the  coefficients  in  the  example, 
(a  -  5)7  .  .  .  a1-  aeb  +  a5b*-  a*b3+  a3b*-  azb5  +  ab6—  b\ 

we  first  place  the  exponent  7  as  a  coefficient  of  the  second 
term.  Then,  to  find  the  coefficient  of  the  third  term,  we 
multiply  7  by  6,  the  exponent  of  a,  and  divide  by  2.  The 
quotient,  21,  is  the  coefficient  of  the  third  terra.  To  find  the 
coefficient  of  the  fourth,  we  multiply  21  by  5,  and  divide 
the  product  by  3 ;  this  gives  35.  To  find  the  coefficient  of 
the  fifth  term,  we  multiply  35  by  4,  and  divide  the  product 
by  4  ;  this  gives  35.  The  coefficient  of  the  sixth  term,  found 

126.  What  is  the  coefficient  of  the  first  terra  ?  What  is  the  coefficient 
of  the  second  term  ?  How  do  you  find  the  coefficient  of  the  third  term  ? 
How  do  you  find  the  coefficient  of  any  term  ?  What  are  the  coefficients 
of  the  first  and  last  terms  ?  How  are  the  coefficients  of  the  exponents 
of  any  two  terms  equally  distant  from  the  two  extremes? 
8 


170  ELEMENTARY      ALGEBKA. 

in  the  same  way,  is  21  ;  that  of  the  seventh,  7  ;  and  that  of 
the  eighth,  1.     Collecting  these  coefficients, 

(a  -  by  = 

-f  21a562—  35a453  +  35a354  —  21a?b5 


NOTE.  —  We  see,  in  examining  this  last  result,  that  th* 
coefficients  of  the  extreme  terms  are  each  1,  and  that  the 
coefficients  of  terms  equally  distant  from  the  extreme  terms 
are  equal.  It  will,  therefore,  be  sufficient  to  find  the  coeffi- 
cients of  the  first  half  of  the  terms,  and  from  these  the 
others  may  be  immediately  written. 

EXAMPLES 

1.  Find  the  fourth  power  of  a  +  b. 

Ans.  a4  +  4a3b  +  6a252  +  4a&3  +  b4. 

2.  Find  the  fourth  power  of  a  —  b. 

Ans.    a*  —  4a3b  +  6a252  —  4a£3  +  b*. 

3.  Find  the  fifth  power  of  a  +  b. 

Ans.   a5  +  5a45  -f  10a352  +  10a2&3  +  5a5*  +  #>. 

4.  Find  the  fifth  power  of  a  —  b. 

Ans.   a5  —  5a4b  +  Wa3b2  —  10a2ft3  +  5ab*  -  b5. 

5.  Find  the  sixth  power  of  a  +  b. 

Gab5  +  bs. 


«8.  Find  the  sixth  power  of  a  —  b. 

a6  —  6a55  +  15a452  —  20a3J3  +  15aW  -  6ab5 


127.  When  the  terms  of  the  binomial  have  coefficients, 
we  may  still  write  out  any  power  of  it  by  means  of  the 
Binomial  Formula. 

X  Let  it  be  required  to  find  the  cube  of  2c  +  3d. 
(a  +  b)3  —  a3  +  Sazb  +  3abz  +  b3. 


POWERS      OF      BINOMIALS.  171 

Here,  2c  takes  the  place  of  a  in  the  formula,  and  3d  the 
place  of  b.     Hence,  we  have, 


.      (I.) 

and  by  performing  the  indicated  operations,  we  have, 
(2c  +  3d)3  =  8c3  +  36c9-J 


If  we  examine  the  second  member  of  Equation  (  1  ),  we 
see  that  each  term  is  made  up  of  three  factors:  1st,  the 
numerical  factor  ;  2d,  some  power  of  2c  ;  and  3d,  some 
power  of  3d.  The  powers  of  2c  are  arranged  in  descend- 
ing order  towards  the  right,  the  last  term  involving  the  0 
power  of  2c  or  1  ;  the  powers  of  3d  are  arranged  in  ascend- 
ing order  from  the  first  term,  where  the  0  power  enters,  to 
the  last  term. 

The  operation  of  raising  a  binomial  involving  coefficients, 
is  most  readily  effected  by  writing  the  three  factors  of  each 
term  in  a  vertical  column,  and  then  performing  the  multipli- 
cations as  indicated  below. 

Find,  by  this  method,  the  cube  of  2c  +  3d. 

OPERATION. 

1+3        +3        +1         Coefficients. 
8c3  -f    4c2     +    2c      +1         Powers  of  2c 
1     +    3d     +    Qd2   +  2ld3    Powers  of  3d 

(2c  +  d)3  =  8c3  +  3Qczd  +  54cc?2  +  27d3 

The  preceding  operation  hardly  requires  explanation.  In 
the  first  line,  write  the  numerical  coefficients  corresponding 
to  the  particular  power  ;  in  the  second  line,  write  the  de- 
scending powers  of  the  leading  term  to  the  0  power  ;  in  the 
third  line,  write  the  ascending  powers  of  the  following  term 
from  the  0  power  upwards.  It  will  be  easiest  to  commence 


172  K  I.  K  M  E  N  T  A  U  Y       ALGEBRA. 

the  second  line  on  the  right  hand.    The  multiplication  should 
be  performed  from  above,  downwards. 

8.  Find  the  4th  power  of  3a2<  —  2bd. 

(a  +  b)<  =  a4  +  4«35  +  6«2&2  +  4a53  +  b*. 

1+4  +0  +4  '+    1 

81a8c4  +    27a6c3      +      9a*c2         +    3a2c          +    1 
1  2bd        +      4b*d2        -    8b3d3        +  16b*d  . 


BlaV  — 


9.  What  is  the  cube  of  3x  -  Gy  ? 
Ans.    27cc3  —  162cc2y 

10.  What  is  the  fourth  power  of  a  —  35? 

Ans.   a*  —  I2a3b  +  54a2i2  —  108«i3  +  Sib4. 


11.  What  is  the  fifth  power  of  c  — 
Ans.   c5  —  lOcV?  +  40c3t?2  —  8Qc"d3  +  BOcd*  —  32f7». 


12.  What  is  the  cube  of  5a  —  3d? 

Ans.    I25a3  —  225a?d  +  I35adz  -  2 


*  This  Ingenious  method  of  writing  the  development  of  a  binomial   Is  due    to 
fr-.ofewor  WILLIAM  G.  PECK,  of  Columbia  College. 


EXTRACTION       OF       ROOTS.  173 


CHAPTER    VH. 

SQUARE   ROOT.    RADICALS   OF  THE  SECOND 
DEGREE. 

12 §.     THE  SQUARE  ROOT  of  a  number  is  one  of  its  two 

equal  factors.  Thus,  6  X  6  =  36;  therefore,  6  is  the  square 
root  of  36. 

The  symbol  for  the  square  root,  is  y' ,  or  the  fractional 
exponent  \ ;  thus, 

r         * 

ya,    or    a  , 

indicates  the  square  root  of  a,  or  that  one  of  the  two  equal 
factors  of  a  is  to  be  found.  The  operation  of  finding  sucli 
factor  is  called,  Extracting  the  Square  Hoot. 

129.     Any  number  which  can  be  resolved  into  two  equal 
integral  factors,  is  called  a  perfect  square. 

The  following  Table,  verified  by  actual  multiplication,  in- 
dicates all  the  perfect  squares  between  1  and  100. 

TABLE. 

1,     4,     9,     16,     25,     36,     49,     64,     81,     100,     squares. 
1,     2,     3,      4,       5,       6,       7,       8,       9,       10,      roots. 

128.  What  is  the  square  root  of  a  number?     Wha    is  the  operation  of 
finding  the  equal  factor  called  ? 

129.  What  is  a  perfect  square  ?     How  many  perfect  squares  are  there 
between  1  and  100,  including  both  numbers  ?    What  arc  they? 


174  ELEMENTARY      ALGEBRA. 

"We  may  employ  this  table  for  finding  the  square  root  of 
any  perfect  square  between  1  and  100. 

Look  for  the  number  in  the  first  line  /  if  it  is  found 
there,  its  square  root  will  be  found  immediately  under  it. 

If  the  given  number  is  less  than  100,  and  not  a  perfect 
square,  it  will  fall  between  two  numbers  of  the  ujyper  line,  and 
its  square  root  will  be  found  between  the  ttco  numbers  directly 
below  ;  the  lesser  of  the  two  will  be  the  entire  part  of  the 
root,  and  will  be  the  true  root  to  within  less  than  1. 

Thus,  if  the  given  number  is  55,  it  is  found  between  the 
perfect  squares  49  and  64,  and  its  root  is  7  and  a  decimal 
fraction. 

NOTE.—  There  are  ten  perfect  squares  between  1  and  100, 
if  -we  include  both  numbers  ;  and  eight,  if  we  exclude  both. 

If  a  number  is  greater  than  100,  its  square  root  will  be 
greater  than  10,  that  is,  it  will  contain  tens  and  units.  Let 
JV  denote  such  a  number,  x  the  tens  of  its  square  root,  and 
y  the  units  ;  then  will, 


XT  =  (x  +  yY  =  a2  +  2a:y  +  t/2  =  «2  +  (2x  +  y)y. 

That  is,  the  number  is  equal  to  the  square  of  the  tens  in  its 
roots,  plus  twice  the  product  of  the  tens  by  the  units,  plus 
the  square  of  the  units. 

EXAMPLE. 

1.  Extract  the  square  root  of  6084. 

Since  this  number  is  composed  of  more  than 
two  places  of  figures,  its  root  will  contain  more  GO  84 

than  one.    But  since  it  is  less  than  10000,  which 
is  the  square  of  100,  the  root  will  contain  but  two  figures  j 
that  is,  units  and  tens. 


SQUARE   ROOT   OF   NUMBERS.       175 

Now,  the  square  of  the  tens  must  be  found  in  the  two 
left-hand  figures,  which  we  will  separate  from  the  other  two 
by  putting  a  point  over  the  place  of  units,  and  a  second  over 
the  place  of  hundreds.  These  parts,  of  two  figures  each, 
are  called  periods.  The  part  60  is  comprised  between  the 
two  squares  49  and  64,  of  which  the  roots  are  7  and  8 ;  hence, 
7  expresses  the  number  of  tens  sought /  and  the  required 
root  is  composed  of  7  tens  and  a  certain  number  of  units. 

The  figure  7  being  found,  we 

write  it  on  the  right  of  the  given  60  84   78 

number,  from  which  we  separate  49 

it  by  a  vertical  line  :  then  we       7  X  2  =  14  8|118~4~ 
subtract  its  square,  49,  from  60,  118  4 

which  leaves  a  remainder  of  11,  0 

to  which  we  biing  down  the  two 

next  figures,  84.  The  result  of  this  operation,  1184,  con- 
tains twice  the  product  of  the  tens  by  the  units,  plus  the 
square  of  the  units. 

But  since  tens  multiplied  by  units  cannot  give  a  product 
of  a  less  unit  than  tens,  it  follows  that  the  last  figure,  4,  can 
form  no  part  of  the  double  product  of  the  tens  by  the  units ; 
this  double  product  is  therefore  found  in  the  part  118,  which 
we  separate  from  the  units'  place,  4. 

Now  if  we  double  the  tens,  which  gives  14,  and  then 
divide  118  by  14,  the  quotient  8  will  express  the  units,  or  a 
number  greater  than  the  units.  This  quotient  can  never  be 
too  small,  since  the  part  118  will  be  at  least  equal  to  twice 
the  product  of  the  tens  by  the  units ;  but  it  may  be  too 
large,  for  the  118,  besides  the  double  product  of  the  tens  by 
the  units,  may  likewise  contain  tens  arising  from  the  square 
of  the  units.  To  ascertain  if  the  quotient  8  expresses  the 
right  number  of  units,  we  write  the  8  on  the  right  of  the  14, 
which  gives  148,  and  then  we  multiply  148  by  8.  This 
multiplication  being  effected,  gives  for  a  product,  1184,  a 


176  ELEMENTARY       ALGEBRA. 

number  equal  to  the  result  of  the  first  operation.  Hav- 
ing subtracted  the  product,  we  find  the  remainder  equal 
to  0;  hence,  78  is  the  root  required.  In  this  operation, 
we  form,  1st,  the  square  of  the  tens ;  2nd,  the  double 
product  ot  the  tens  by  the  units ;  and  3d,  the  square  of 
the  units. 

Indeed,  in  the  operations,  we  have  merely  subtracted  from 
the  given  number  GO 84  :  1st,  the  square  of  7  tens,  or  of  70 ; 
2d,  t\vice  the  product  of  70  by  8 ;  and,  3d,  the  square  of  8 ; 
that  is,  the  three  parts  which  enter  into  the  composition  of 
the  square,  70  -f  8,  or  78  ;  and  since  the  result  of  the  sub- 
traction is  0,  it  follows  that  78  is  the  square  root  of  6084. 

ISO.  The  operations  in  the  last  example  have  been  per- 
formed on  but  two  periods,  but  it  is  plain  that  the  same 
methods  of  reasoning  are  equally  applicable  to  larger  num- 
bers, for  by  changing  the  order  of.  the  units,  AVC  do  not 
change  the  relation  in  which  they  stand  to  each  other. 

Thus,  in  the  number  60  84  95,  the  two  periods  60  84, 
havo  the  same  relation  to  each  other  as  in  the  number 
60  84  ;  and  hence  the  methods  used  in  the  last  example  are 
equally  applicable  to  larger  numbers. 

131.  Hence,  for  the  ex-traction  of  the  square  root  of 
numbers,  we  have  the  following 

E  u  L  E  . 

I.  Point  off  the  given  number  into  periods  of  two  figures 
each,  beginning  at  tlie  right  hand:  . 

H.  Note  the  greatest  perfect  square  in  the  first  period  on 
tJie  left,  andpkice  its  root  on  the  right,  after  the  manner  of 

131.  Give  the  rule  for  the  extraction  of  the  square  root  of  numbers? 
What  is  the  first  step  ?  What  the  second  ?  What  the  third  ?  What  the 
fourth?  What  the  filth? 


SQUARE      KOOT      OF      NUMBERS.  177 

a  quotient  in  division  y  then  subtract  the  square  of  this 
root  from  the  first  period,  and  bring  down  the  second  period 
for  a  remainder: 

III.  Double  the  root  already  found,  and  place  the  result 
on  the  left  for  a  divisor.     Seek  hoic  many  times  the  divisor 
is  contained  in  the  remainder,  exclusive  of  the  right-hand 
figure,  and  place  the  figure  in  the  root  and  also  at  the  right 
of  the  divisor  : 

IV.  Multiply  the  divisor,  thus  augmented,  by  the  last 
figure  of  the  root,  and  subtract  the  product  from  the  re- 
mainder, and  bring  down  the  next  period  for  a  new  remain- 
der.    J3ut  if  any  of  the  products  should  be  greater  than 
the  remainder,  diminish  the  last  figure  of  the  root  by  one  : 

V.  Double  the  ichole  root  already  found,  for  a  new  di- 
visor, and  continue  the  operation  as  before,  until  all  the 
periods  are  brought  down. 

132.  NOTE. — 1.  If,  after  all  the  periods  are  brought 
down,  there  is  no  remainder,  the  given  number  is  a  perfect 
square. 

2.  The  number  of  places  of  figures  iu  the  root  will  always 
be  equal  to  the  number  of  periods  into  which  the  given 
number  is  divided. 

3.  If  the  given  number  has  not  an  exact  root,  there  will 
be  a  remainder  after  all  the  periods  are  brought  down,  in 
which  case  ciphers  may  be  annexed,  forming  new  periods, 
for  each  of  which  there  will  be  one  decimal  place  in  the  root. 

132.  What  takes  place  v.-hen  the  given  number  id  a  perfect  square ! 
How  many  places  of  figures  will  there  be  in  the  root?  If  the  given  num- 
ber is  not  a  perfect  square,  what  may  be  done  after  all  the  periods  are 
brought  dowii  ? 


178 


ELEMENTARY    ALGEBKA 


EXAMPLES. 
1.  What  is  the  square  root  of  36729  ? 


In  this  example  there  are 
two  periods  of  decimals, 
and,  hence,  two  places  of 
decimals  in  the  root. 


3  67  29    191.64  + 
1 


29 


261 


381 


3826 


629 
381 


24800 
22956 


38324 


31104  Rem. 

2.  To  find  the  square  root  of  7225.  Ans.  85. 

3.  To  find  the  square  root  of  17689.  Ans.  133. 

4.  To  find  the  square  root  of  994009.  Ans.  997. 

5.  To  find  the  square  root  of  85673536.  Ans.  9256. 

6.  To  find  the  square  root  of  67798756.  Ans.  8234. 

7.  To  find  the  square  root  of  978121.  Ans.  989. 

8.  To  find  the  square  root  of  956484.  Ans.  978. 

9.  What  is  the  square  root  of  36372961  ?  -    Ans.  6031. 

10.  What  is  the  square  root  of  22071204  ?       Ans.  4698. 

11.  What  is  the  square  root  of  106929?  Ans.  327. 

12.  What  of  12088868379025?  Ans.  3476905. 

13.  What  of  2268741  ?  Ans.  1506.23  +. 

14.  What  of  7596796?  Ans.  2756.22  +, 

15.  What  is  the  square  root  of  96  ?        Ans.  9.79795  +. 

16.  What  is  the  square  root  of  153?    Ans.  12.36931  -f . 

17.  What  is  the  square  root  of  101  ?    Ans.  10.04987  -K 


SQUARE      ROOT     OF     FRACTIONS.  179 

18.  What  of  285970396644  ?  Ans.  534762. 

19.  What  of  41605800625  ?  Ans.  203975. 

20.  What  of  48303584206084?  Ans.  6950078. 

EKTRACTION   OF  THE  SQUARE  ROOT   OF  FRACTIONS. 

133.  Since  the  square  or  second  power  of  a  fraction  is 
obtained  by  squaring  the  numerator  and  denominator  sepa- 
rately, it  follows  that  „ 

The  square  root  of  a  fraction  will  be  equal  to  the  square 
root  of  the  numerator  divided  by  the  square  root  of  tJie 
denominator. 

For  example,  the  square  root  of   7-5-    is  equal  to   r:  for, 
a       a  __  a^ 

V        ^       T      TO    * 

b        b       bz 
1.  What  is  the  square  root  of  -?  Ans.  -• 


2. 

What  is  the  square  root 

of 

16' 

Ans.   -  - 
4 

8. 

What  is  the 

square  root 

of 

81 

8 
Ans.   -• 

4. 

What  is  the 

square  root 

of 

256 
361 

16 
Ans.  -. 

5. 
6. 

What  is  the 
What  is  the 

square  root 
square  root 

of 
of 

64* 
4096 

Ans.   I. 
64 

A<w  o 

61009 

•   247 

7. 

What  is  the 

square  root 

of 

582169  ? 

763 

A  -JT  « 

956484  ' 

*    978 

183.  To  what  is  the  square  root  of  a  fraction  equal  ? 


ELEMENTAKY        ALGEBKA. 

134.  If  the  numerator  and  denominator  are  not  perfect 
squares,  the  root  of  the  fraction  cannot  be  exactly  found. 
We  can,  however,  easily  find  the  approximate  root. 

RULE. 

Multiply  both  terms  of  tlie  fraction  by  the  denominator  : 
Then  extract  the  square  root  of  the  numerator,  and  divide 
this  root  by  the  root  of  the  denominator  ;  the  quotient  will 
be  the  approximate  root. 

g 

1.  Find  the  square  root  of  -  • 

5 

Multiplying  the  numerator  and  denominator  by  5 

\/f  -  v/i = 

hence,     (3.8729  -f )  -f-  5  =     .7745  -f   =  Ana. 

7 

2.  What  is  the  square  root  of  -  ?  Ans.   1.32287  f. 

14 

3.  What  is  the  square  root  of  —  ?         Ans.   1.24721  +. 

9 

4.  What  is  the  square  root  of   11—?     Ans.   3.41869  +. 

16 

13 

5.  What  is  the  square  root  of   7—?       Ans.   2.71313  +. 

36 

6.  What  is  the  square  root  of   8—?       Ans.   2.88203  -f. 

g 

7.  What  is  the  square  root  of   —  ?         Ans.   0.64549  -f . 

Q 

8.  What  is  the  square  root  of    10 — ?     Ans.   3.20936  +. 

134.  What  is  the  rule  when  the  numerator  and  denominator  are  not 
perfect  squares  ? 


SQUARE  ROOT  OF  MONOMIALS.       181 

135.  Finally,  instead  of  the  last  method,  we  may,  if  wo 
please, 

Change  the  common  fraction  into  a  decimal,  and  continue 
tho>  division  until  the  number  of  decimal  places  is  double 
the  number  of  places  required  in  the  root.  TJien  extract 
the  root  of  the  decimal  by  the  last  rule. 

EXAMPLES. 

1.  Extract  the  square  of   —    to  within   .001.     This  num- 
ber, reduced  to  decimals,  is  0.785714  to  within  0.000001 ;  but 
the  root  of  0.785714  to  the  nearest  unit,  is   .886;  hence, 

0.886  is  the  root  of   —    to  within  .001. 
14 

/  1  3 

2.  Find  the    \/2~    to  within  0.0001.     Ans.    1.6931  +. 

3.  What  is  the  square  root  of   —  ?         Ans.   0.24253  +. 

7 

4.  What  is  the  square  root  of   -?  Ans.    0.93541  +. 

8 

g 

5.  What  is  the  square  root  of   -  ?  Ans.    1.29099  +. 

3 


EXTRACTION   OF  THE  SQUARE   ROOT   OF   MONOMIALS. 

136.  In  order  to  discover  the  process  for  extracting  the 
square  root  of  a  monomial,  we  must  see  how  its  square  ia 
formed. 

By  the  rule  for  the  multiplication  of  monomials  (Art.  42), 
we  have, 

x  5a2£3c  =  25a4i6c2  ; 


135.  What  is  a  second  method  of  finding  the  approximate  root? 

136.  Give  the  rule  for  extracting  the  square  root  of  monomials? 


182  ELEMENTARY       ALGEBRA. 

that  is,  in  order  to  square  a  monomial,  it  is  necessary  to- 
square  its  coefficient  and  double  the  exponent  of  each  oftht 
letters.  Hence,  to  find  the  square  root  of  a  monomial,  we 
have  the  following 

RULE. 

I.  Extract  the  square  root  of  the  coefficient  for  a  new 
coefficient : 

H.  Divide  tJie  exponent  of  each  letter  by  2,  and  then 
annex  all  the  letters  with  their  new  exponents. 

Since  like  signs  in  two  factors  give  a  plus  sign  in  the  pro- 
duct, the  square  of  —  a,  as  well  as  that  of  -f  a,  will  be 
+  a2;  hence,  the  square  root  of  a2  is  either  +  «»  01 
—  a.  Also,  the  square  root  of  25«254,  is  either  +  5aJ2, 
or  —  5abz.  Whence  we  conclude,  that  if  a  monomial  is 
positive,  its  square  root  may  be  affected  either  with  the  sign 
+  or  —  ;  thus,  1/9O4  =  ±  3a2 ;  for,  +  3a2  or  —  3a2, 
squared,  gives  -'-  9a4.  The  double  sign  ± ,  with  which  the 
root  is  affected,  is  read  plus  and  minus. 

EXAMPLES. 

1.  What  is  the  square  root  of   64a6&4? 

and,  -y/64a664  =  —  8a352;  for  —  8a352x  —  8a352= 
Hence,  -y/64a664  =   ±  Sa3b*. 

2.  Find  the  square  root  of  625a258c6.  ±  25a£4c3. 

3.  Find  the  square  root  of  576a456c8.  ±  24a2£3c4. 

4.  Find  the  square  root  of  196a;6y2s4.  ±  14x3yz2. 

5.  Find  the  square  root  of  441a8£6c10J16.  ±  21a453c5J8. 

6.  Find  the  square  root  of  784a12&14c16d'2.  ±  2Saeb'ced. 

7.  Find  the  square  root  of  8la*b*c*.  ±  9o4£2c3. 


IMPERFECT      8  Q  \I  A  K  E  S  .  1 83 

NOTES. — 137.  1.  From  the  preceding  rtuo  it  follows, 
that  when  a  monomial  is  a  perfect  square,  its  numerical 
coefficient  is  a  perfect  square,  arid  all  its  exponents  even 
numbers.  Tims,  25«4&2  is  a  perfect  square. 

2.  If  the  proposed  monomial  were  negative,  it  would  be 
impossible  to  extract  its  square  root,  since  it  has  just  been 
shown  (Art.  136)  that  the  square  of  every  quantity,  whether 
positive  or  negative,  is  essentially  positive.  Therefore, 


are  algebraic  symbols  which  indicate  operations  that  cannot 
be  performed.  They  are  called  imaginary  quantities,  or 
rather,  imaginary  expressions,  and  are  frequently  met  with 
in  the  resolution  of  equations  of  the  second  degree. 

IMPERFECT   SQUARES. 

13§.  When  the  coefficient  is  not  a  perfect  square,  or 
when  the  exponent  of  any  letter  is  uneven,  the  monomial  is 
an  imperfect  square :  thus,  QSab*  is  an  imperfect  square. 
Its  root  is  then  indifcated  by  means  of  the  'adical  sign ;  thus. 


Such  quantities  are  called,  radical  quantities,  or  radicals  of 
the  second  degree :  hence, 

A  RADICAL  QUANTITY,  is  the  indicated  root  of  an  imperfect 
power. 

137.  When  is  a  monomial  a  perfect  square  ?    What  monomials  are 
these  whose  square  roots  cannot  be  extracted  ?    What  are  such  expres- 
sions called? 

138.  When  is  a  monomial  an  imperfect  square  ?     What  are  such 
'Jtios  called  ?     What  is  A  radical  quantity  ? 


184  ELEMENTARY      ALGEBRA. 

TRANSFORMATION      OF      RADICALS. 

139.  Let  a  and  b  denote   any  two  numbers,  and  J5 
the  product  of  their  square  roots:  then, 

V^  X  V*  =  P CD 

Squaring  both  members,  we  have, 

a  X  b  =  p*      ....     (2.) 

Then,  extracting  the  square  root  of  both  members  of  (2), 

i/ao  =  p (3.) 

And  since  the  second  members  are  the  same  in  Equations 
( 1 )  and  (  3 ),  the  first  members  are  equal :  that  is, 

The  square  root  of  the  product  oftico  quantities  is  equal 
to  the  product  of  their  square  roots. 

140.  Let  a  and  b  denote   any  two  numbers,  and   q 
the  quotient  of  their  square  roots ;  then, 

^   =   q (1.) 

Squaring  both  members,  we  have, 

a  .    . 

•=•  =  q2 (2.) 

o 

then  extracting  the  square  root  of  both  members  of  (2), 


..... 

and  since  the  second  members  are  the  same  in  Equations  (  1  ) 
and  (3  ),  the  first  members  are  equal  ;  that  is, 

139.  To  what  is  the  square  root  of  the  product  of  two  quantities  equal? 

140.  To  what  is  the  square  root  of  the  quotient  of  two  quau  tiling 
ual? 


equal? 


TRANSFORMATION       CF      RADICALS.       185 

The,  square  root  of  the  quotient  of  two  quantities  is  equal 
to  the  quotient  of  their  square  roots. 

These  principles  enable  us  to  transform  radical  expres- 
sions, or  to  reduce  them  to  simpler  forms ;  thus,  the  expres- 
sion, 

98«&*  =  495*  X  2a ; 


hence,  yOSab*  =    -v/49i4  x  2a; 

and  by  the  principle  of  (Art.  139), 


x  2a  = 
In  like  manner, 


X  5bd       = 
•V/864a2£5cn  =   -v/144a264c10  x  Qbc  —  12ab*c5 

The  COEFFICIENT  of  a  radical  is  the  quantity  without  the 
sign  ;  thus,  in  the  expressions, 


the   quantities    752,    3abc,    12a52c5,    are  coefficients  of  the 
radicals. 

141.  Hence,  to  simplify  a  radical  of  the  second  degree, 
we  have  the  following 

RULE. 

I.  Divide  the  expression  under  the  radical  sign  into  two 
factors,  one  of  which  shall  be  a  perfect  square  : 

II.  ^Extract  the  square  root  of  the  perfect  square,  and 
then  multiply  this  root  by  the  indicated  square  root  of  the, 
remaining  factor. 

141.  Give  the  rule  for  simplifying  radicals  of  the  second  degree.  How 
do  you  determine  whether  a  given  number  has  a  factor  which  is  a  perfect 
square  ? 


186  ELEMENTAKY       A  L  G  K  B  E  A  . 


-To  determine  if  a  given  number  has  any  factor 
which  is  a  perfect  square,  we  examine  and  see  if  it  is  divi- 
sible by  either  of  the  perfect  squares, 

4,     9,     1C,     25,     36,     49,     64,     81,  &c. ; 

if  it  is  not,  we  conclude  that  it  does  not  contain  a  factor 
•vhich  is  a  perfect  square. 

EXAMPLES. 

Reduce  the  following  radicals  to  their  simplest  form : 

Ans.  5a-\/3abc. 


2.   Vl2Sb5a6dz.  Ans. 


3*  /  QO/797j8/»  A  ,vj  o 

•     \t  O  —  C(,  U  C«  -/A/co. 


4.  i/256a2&4c8.  Ans. 


5.  -v/1024a967c*.  Ans.  32alb3c2yabc. 

6. 


7.  \/675a755c2<?.  Ans. 

Ans. 


9.  V1008a9tf7ra8.  Ans. 


10.  V215Ga10Z»8c6. 
11. 


142.  NOTES. — 1.  A  coefficient,  or  a  factor  of  a  coeffi- 
cient, may  be  carried  under  the  radical  sign,  by  squaring  it. 
Thus, 


2.  2abid    =  2 


142.  How  may  a  coefBcient  or  factor  be  carried  under  the  radical  sign 
To  what  is  the  square  root  of  a  negative  quantity  equal  ? 


ADDITION       OF      RADICAL 


187 


3. 
4. 


c2). 


2.  The  square  root  of  a  negative  quantity  may  also 
simplified;  thus, 


x   -  1    =    -y/9  X   -y/^1    —   3-/^~l, 
^  4a2  —   -\/4a?  x  •/—  1   —  Say'—  1  ;       also, 

X 


and, 


that  is,  the  square  root  of  a  negative  quantity  is  equal  to 
the  square  root  of  the  same  quantity  with  a  positive  sign, 
multiplied  into  the  square  root  of  —  1. 

Reduce  the  following : 


2.   V—  128«465. 
3. 


Ans.  6a2 J3c3 


4.      '-  48a3£c5. 


ADDITION    OF    EADICALS. 

143.  SIMILAR  RADICALS,  of  the  second  degree,  are  those 
in  which  the  quantities  under  the  sign  are  the  same.  Thus, 
the  radicals  3-y/7>,  and  5c^/b  are  similar,  and  so  also  are 
2,  and 


144.     Radicals  are  added  like  other  algebraic  quantities  ; 

hence,  the  following 


143.  \V  h;;t  are  similar  radicals  of  the  second  degree  ? 

144.  Give  the  rule  for  the  addition  of  radicals  of  the  second  degree  ? 


188  ELEMENTARY      ALGEBKA 


BULE. 

L  If  the  radicals  are  similar,  add  their  coefficients,  and 
to  the  sum  annex  the  common  radical  : 

II.  If  the  radicals  are  not  similar,  connect  them  together 
icith  their  proper  signs. 


Thus,         3ay^  +  Sc-y/J  =   (3a  4- 
In  like  manner, 

li/2a  +  3-/2a  =  (7  +  3)-v/2 

NOTES.  —  1.  Two  radicals,  which  do  not  appear  to  be  sim- 
ilar at  first  sight,  may  become  so  by  transformation  (Art. 
141.) 

For  example, 


2-V/45  +  3-v/5    =   6-v/S  +  3-/5   =   9^5- 

2.  When  the  radicals  are  not  similar,  the  addition  or  sub- 
traction can  only  be  indicated.  Thus,  La  order  to  add  3  -y/J 
to  5-v/a,  we  write, 

5-v/a  +  3-/J. 

Add  together  the  following  : 


2.   -/SOo^2    and    yY2a4Z>2. 


3.        -  -    and 
5 


7^5 

\/  —  • 
V   15 


4.   -y/125   and    v^OOo2.  ^/w.  (5  +  10a)-v/5 


/ 

V 


oo  10 

294 


11. 


SUBTRACTION       OF       RADICALS.  189 

'9Sazx   and 


Ans.  1a^/2x  +  6  i/x2  —  a2. 


12.  -v243    and    10-/363.  Ans. 

13.  -v/320o2i2   and    -\/245a8b6.     Ans.  (Sab  +  7a*53)  i/5. 

14.  y^oa6*7   and    -v/300a665.  Ans.  (5a353 


SUBTRACTION    OF    RADICALS. 

145.     Radicals  are  subtracted  like  other  algebraic  quan- 
tities ;  hence,  the  following 

'RULE. 

I.  If  the  radicals  are  similar,  subtract  the  coefficient  of 
the  subtrahend  from  that  of  the  minuend,  and  to  tJie  differ- 
ence annex  the  common  radical  : 

II.  If  the  radicals  are  not  similar,  indicate  the  operation 
by  the  minus  sign. 


EXAMPLES. 

1.  What  is  the  difference  between  3a-\/b  and  a 
Here,  3a\/b  —  a^fb  =  1a-\/b.     Ans. 


14  K.  Give  the  rule  for  the  subtraction  of  radicals. 


190  ELEMENTARY      ALGEBRA. 

2.  From   Oav/2762  subtract   6ay2752. 
First,     Qa-^27b2  —  27a£-y/3,   and  Ga-v/2762  =   1 
and,  27aJy/3  —  l8ab\/3  —  9a6-v/3.    ^4.ns. 

Find  the  differences  between  the  following : 
3.   -1/75   and 


4. 


and 


.  (2ab  — 


27 
and 


j_ 
45 


363   and 
and 


6. 
7. 
8. 

**\     and 

10.  -/32002   and    -SOa2. 


Ans. 


.  4a-/5". 


12. 


13.  -v/ll2a866   and    -v/28a866. 


MULTIPLICATION    OF    RADICALS. 

146.    Radicals  are  multiplied  like  other  algebraic  quan- 
tities ;  hence,  we  have  the  following 

EULE. 
L  Multiply  the  coefficients  together  for  a  new  coefficient: 

146.  Give  the  rule  for  the  multiplication  of  radicals. 


DIVISION       OF       KADICAL8-.  391 

II.  Multiply  together  the  quantities  under  the  radical 
signs: 
HI.  Then  reduce  the  result  to  its  simplest  form. 

1.  Multiply   3a\/bc  by  2\/ab. 

3a^/bc  X  2<\/ab  =  3a  X  2  x  -\/bc  x   ^/ab. 
which,  by  Art.  139,  =  6a\/b2ac  =  §dbi/ac. 

Multiply  the  following  : 

2.  3\/5ab  and  4y^0a.  Ans.  120«v^- 

3.  2a-[/bc  aad   Sa^bc.  Ans.  6a25c. 

4.  Za^/aT+W  and   —  3a<^a?~+&.   A.  - 

5.  Zabi/a  +  6  and 

6.  3-V/2   and   2-v/8.  ^ws.  24. 

7.  fy^«^  and  TSO 

8.  2x  +  -\/b  and   2«  —  ^/b.  Ans.  4cc2  —  b. 

9. 


10.  3«-/27a3  by 


DIVISION     OF     EADICALS. 


147.     Radical  quantities  are  divided  like  other  algebraic 
quantities  ;  hence,  we  have  the  following 


RULE. 


I.   Divide  the  coefficient  of  the  dividend  by  the  coefficient 
of  the  divisor,  for  a  new  coefficient  : 


147.  Give  the  rule  for  the  division  of  radicals. 


192 


ELEMENTARY      ALGEBRA. 


II.  Divide  the  quantities  under  the  radicals,  in  the  same 
manner  : 
HI.    Then  reduce  the  result  to  its  simplest  form. 


1.  Divide 


EXAMPLES. 

by  < 


' 


—  '  =  2,    new  coefficient. 


Art.  140, 


hence,  the  quotient  is     2    X  -  =  - 


2.  Divide 


by    25-y/c. 


Ans.       V/-- 


3.  Divide  12ac-y/6Jc    by 

4.  Divide  6a\X%6*    by 

5.  Divide  4a2-/5065    by 

6.  Divide  26a35-/81a2&2    by    13a\/Qab.    A. 
1.  Divide  84a354v^7ac    by    42a5-v/3ff-      ^L 

8.  Divide  -v/i^   ^7    V^- 

9.  Divide  6a2J2-/20a3    by    12-v/5«. 

10.  Divide  6a-v/10i2    by    3-v/sl 

11.  Divide  485*  -/15    by    2J2y^ 

12.  Divide  Sa2^3-/^3    by    2a-v/28d 

13.  Divide  96a4c3/98i5    by 


4a5v/3". 


a3i2. 


360R 
.  2ai*c3J. 
.  14a3ic2. 


SQUARE   BOOT   OF   POLYNOMIALS.     193 


14.  Divide    27a566-/2«"3    by    ^Ja.       Ans. 

15.  Divide    18aBb6^/8a*    by    6ab</a?.      Ans. 

SQUARE     ROOT     OF     POLYNOMIALS. 

148.  Before  explaining  the  rule  for  the  extraction  of  the 
square  root  of  a  polynomial,  let  us  first  examine  the  squares 
of  several  polynomials  :  we  have, 

(a  +  b)2  =  a?  +  2ab  +  b\ 

(a  +  b  +  c)2  =  a2  +  lab  +  b2  +  2(a  +  fyc  +  c2, 

(a  -f-  b  +  c  +  (I)*  =  a2  +  2ai  +  b2  +  2(a  +  %  +  c3 

+  2(a  +  J  +  c)c?  -f-  f?2. 

The  /ate  by  which  these  squares  are  formed  can  be  enun 
ciated  thus  : 

The  square  of  any  polynomial  is  equal  to  the  square  of 
the  first  term,  plus  twice  the  product  of  the  first  term  by  the 
second,  plus  the  square  of  the.  second;  plus  twice  the  first 
two  terms  multiplied  by  the  third,  plus  the  square  of  the 
third  ;  plus  twice  the  first  three  terms  multiplied  by  the 
fourth,  plus  the  square  of  the  fourth  ;  and  so  on. 

149.  Hence,  to  extract  the  square  root  of  a  polynomial, 
we  have  the  following 

RULE. 

I.  Arrange  the  polynomial  \oith  reference  to  one  of  its 
letters,  and  extract  the  square  root  of  tJie  first  term  :  thi* 
witt  give  the  first  term  of  the  root  : 

148.  What  is  the  square  of  a  binomial  equal  to?    What,  is  the  square 
of  a  trinomial  equal  to  ?    To  what  is  the  square  of  any  polynomial  equal  ? 

149.  Give  the  jule  for  extracting  the  square  root  of  a  polynomial  ? 
What  is  the  first  step?    Wl  at  the  second  ?    What  the  third  ?    What  the 
fourth  ? 

9 


194;  KLKMKNTAKY       ALGEBEA. 

II.  Divide  the  second  term  of  the  polynomial  by  double 
the  first  term  of  the  root,  and  the  quotient  will  be  the  second 
term  of  the  root : 

HI.  Then  form  the  square  of  the  algebraic  sum  of  the 
two  terms  of  the  root  found,  and  subtract  it  from  the  first 
polynomial,  and  then  divide  the  first  term  of  the  remainder 
by  double  the  first  term  of  the  root,  and  the  quotient  will  be 
the  third  term  : 

IV.  Form  the  double  product  of  the  sum  of  the  first  and 
second  terms  by  the  third,  and  add  the  square  of  the  third  / 
then  subtract  this  result  from  the  last  remainder,  and  divide 
the  first  term  vf  the  result  so  obtained,  by  double  the -first 
term  of  the  root,  and  the  quotient  will  be  the  fourth  term. 
Then  proceed  in  a  similar  manner  to  find  the  other  terms. 

EXAMPLES. 

1.  Extract  the  square  root  of  the  polynomial, 

49«252  —  24a£3  +  25a*  —  30a3b 
First  arrane  it  with  reference  to  the  letter  a. 


165' 


25a4  —  3Qa3b 


5a2  —  Sab 


10a2 


IQb*      .     .     1st  Rem. 


0     . 2c?  Hem. 

After  having  arranged  the  polynomial  with  reference  to 
a,  extract  the  square  root  of  25«4 ;  this  gives  5a2,  which 
is  placed  at  the  right  of  the  polynomial :  then  divide  the 
second  term,  -  SQa3b,  by  the  double  of  5 a2,  or  lOa2; 
the  quotient  is  —  Sab,  which  is  placed  at  the  right  of  5a2. 
Hence,  the  first  two  terms  of  the  root  are  5 a2  —  Sab. 
Squaring  this  binomial,  it  becomes  25a4  —  30«3J  +  Qa2b2, 
which,  subtracted  from  the  proposed  polynomial,  gives  a 
remainder,  of  which  the  first  term  is  40a2ia.  Dividing  thia 


'SQUARE      ROOT      OF      POLYNOMIALS.         195 

first  term  by  10a2,  (the  double  of  5«2),  the  quotient  is 
-f-  4b2  ;  this  is  the  third  term  of  the  root,  and  is  written  on 
the  right  of  the  first  two  terms.  By  forming  the  double 
product  of  5a2  —  Sab  by  4£>2,  squaring  4£»2,  and  taking 
the  sum,  we  find  the  polynomial  40a2£2  —  24ab3  +  16&4, 
which,  subtracted  from  the  first  remainder,  gives  0.  There- 
fore, 5a2  —  Sab  +  4£2  is  the  required  root. 

2.  Find  the  square  root  of  a*+  4a3«+6a2a;2+4cfa^+  cc4* 

-4ns.   «2+  2a«  +  a2. 

3.  Find  the  square  root  of   a4—  4a3a+6a2cc2—  4a#3+  JB*. 

Ans.   a2  —  2ax  +  «2. 

4.  Find  the  square  root  of 

4cc6  +  12x5  +  So4  -  2x3  +  Va:2  —  2aj  +  1. 

Ans.   2x?  +  3a;2  —  x  +  1. 

5.  Find  the  square  root  of 


3a2  —  2ab  +  4i2. 

6.  What  is  the  square  root  of 

a4  —  4ax3  +  4a2a2  —  4a2  +  8aa;  +  4  ? 

-4;i5.   a2  —  2aa  —  2. 

7.  What  is  the  square  root  of 

9«2  —  12aj  +  6xy  +  yz  —  4y  +  4  ? 

-4ws.  3a;  +  y  —  2. 

8.  What  is  the  square  root  of  y4  —  2y2a;2  +  2x2  —  2y2 
1  +  a*  ?  ^tn5.  y2  —  &  —  1. 

9.  What  is  the  square  root  of   9a4£4  —  30as£3+  25a252? 

Ans.   3a2b2  —  Sab. 
10.  Find  the  square  root  of 

48a52c3  f  3652c* 
-  36a25c3  +  9a*c2. 
Ans.   5a?b  -  3«2c  —  4abc  -f 


196  ELEMENTARY      ALGEBKA. 

I5O.  Wo  will  conclude  this  subject  with  the  following 
remarks  : 

1st.  A  binomial  can  never  be  a  perfect  square,  since  we 
know  that  the  square  of  the  most  simple  polynomial,  viz., 
a  binomial,  contains  three  distinct  parts,  which  cannot  ex- 
perience any  reduction  amongst  themselves.  Thus,  the 
expression  a2  -f  bz,  is  not  a  perfect  square  ;  it  wants  the 
term  ±  2a£,  in  order  that  it  should  be  the  square  of  a  ±  b. 

2d.  In  order  that  a  trinomial,  when  arranged,  may  be  a 
perfect  square,  its  two  extreme  -terms  must  be  squares,  and 
the  middle  term  must  be  the  double  product  of  the  square 
roots  of  the  two  others.  Therefore,  to  obtain  the  square 
root  of  a  trinomial  when  it  is  a  perfect  square  :  Extract  the 
roots  of  the  two  extreme  terms,  and  give  these  roots  the  same 
or  contrary  signs,  according  as  the  middle  term  is  positive 
or  negative.  To  verify  it,  see  if  the  double  product  of  the 
two  roots  is  the  same  as  the  middle  term  of  the  trinomial. 
Thus, 

9«6  —  48a4£2  +  64a2J4,    is  a  perfect  square, 


since,        y^6  =  3c/3,    and    -/64«2i4  —   —  Sab2  ; 
and  also, 

2  x  3a3  X  —  Sab2  =   —  4Sa*b2  =  the  middle  term. 

But,  4a2  4-  14a#  +  952  is  not  a  perfect  square  :  for, 
although  4a2  and  +  9bz  are  the  squares  of  2a  and  3#, 
yet  2  x  2a  X  35  is  not  equal  to  I4ab. 

3d.  In  the  series  of  operations  required  by  the  general 
rule,  when  the  first  term  of  one  of  the  remainders  is  not 
exactly  divisible  by  twice  the  first  term  of  the  root,  we  may 

150.  Can  a  binomial  ever  be  a  perfect  power?  Why  not?  When  is 
a  trinomial  a  perfect  square  ?  When,  in  extracting  the  square  root,  we 
find  that  the  first  term  of  the  remainder  is  not  divisible  by  twice  the  root, 
is  the  polynomial  a  perfect  power  or  not? 


SQUARE      ROOT      OK      PCLYNOMIAI,  S.       197 

conclude  that  the  proposed  polynomial  is  not  a  perfect 
square.  This  is  an  evident  consequence  of  the  course  of 
reasoning  by  which  we  have  arrived  at  the  general  rule  for 
extracting  the  square  root. 

4th.  When  the  polynomial  is  not  a  perfect  square,  it  may 
sometimes  be  simplified  (See  Art.  139). 


Take,  for  example,  the  expression,  <^d*b  +  4a252  -f-  4ab3. 

The  quantity  under  the  radical  is  not  a  perfect  square ; 
but  it  can  be  put  under  the  form  ab(a?  +  4a5  +  4&2.) 
Now,  the  factor  within  the  parenthesis  is  evidently  the 
square  of  a  -f  25,  whence,  we  may  conclude  that, 


+  4a252  +  4a63  =   (a  +  25) 

2.  Reduce    -y/2a26  —  4a52  +  253    to  its  simplest  form. 

Ans.  (a  —  b) 


1#S  ELEMENTARY      ALGEBRA 


CHAPTER    VILL 

EQUATIONS        OF        THE       SECOND        DEGREE. 
EQUATIONS    CONTAINING     ONE    UNKNOWN    QUANTITY. 

151.  AN  EQUATION  of  the  second  degree  containing  bu^ 
one  unknown  quantity,  is  one  in  which  the  greatest  exponent: 
is  equal  to  2.     Thus, 

a2  =  a,      ax2  +  bx  =  c, 
are  equations  of  the  second  degree. 

152.  Let  us  see  to  what  form  every  equation  of  the 
second  degree  may  be  reduced. 

Take  any  equation  of  the  second  degree,  as, 

(1  +«)>-?*-  10   =  5  -?  +  £-     . 

Clearing  of  fractions,  and  performing  indicated  operations, 
we  have, 

4  +  8x  +  4«2  —  3x  —  40  =  20  —  «  +  2x\ 

Transposing  the  unknown  terms  to  the  first  member,  the 
known  terms  to  the  second,  and  arranging  with  reference  to 
the  powers  of  #,  we  have, 

4a2  —  2a;2  +  8x  —  3x  +  x  =  20  +  40  —  4 ; 

151.  What  is  an  equation  of  the  second  degree  ?     Give  an  example. 

152.  To  what  form  may  every  equation  of  the  second  degree  be  reduced? 


EQUATIONS     OF     THE     SECOND     DEGREE. 

and,  by  reducing, 

2x2  +  6x  =  56  ; 

dividing  by  the  coefficient  of  #2,  we  have, 
x2  +  3x  =  28. 

If  we  denote  the  coefficient  of  x  by  2p,  and  the  second 
member  by  q,  we  have, 

xz  +  2px  =  q. 
This  is  called  the  reduced  equation. 

153.  When  the  reduced  equation  is  of  this  form,  it  con- 
tains three  terms,  and  is  called  a  complete  equation.  The 
terms  are, 

FIRST  TERM. — The  second  power  of  the  unknown  quan- 
tity, with  a  plus  sign. 

SECOND  TERM. — The  first  power  of  the  unknown  quantity, 
Math  a  coefficient. 

THIRD  TERM. — A  known  term,  in  the  second  member. 

Every  equation  of  the  second  degree  may  be  reduced  to 
this  form,  by  the  following 

BULK. 

I.  Clear  the  equation  of  fractions,  and  perform  all  the 
indicated  operations  : 

n.  Transpose  all  the  unknown  terms  to  the  first  member, 
and  all  the  known  terms  to  the  second  member  : 

153.  How  many  terms  are  there  in  a  complete  equation  ?  What  is  the 
first  term  ?  "What  is  the  second  term  ?  What  is  the  third  term  ?  How 
many  operations  are  there  in  reducing  an  equation  of  the  second  degree 
to  the  required  form  ?  What  is  the  first  ?  What  the  second  ?  What  the 
third  ?  What  the  fourth  ? 


200  ELEMENTAKY       ALGEBKA. 

TTT.  Reduce  all  the  terms  containing  the  square  of  the 
unknown  quantity  to  a  single  term,  one  factor  of  which  is 
the  square  of  the  unknown  quantity  ;  reduce,  also,  all  thd 
terms  containing  the  first  power  of  the  unknown  quantity, 
to  a  single  term  : 

IV.  Divide  both  members  of  the  resulting  equation  by 
the  coefficient  of  the  square  of  tJie  unknown  quantity. 

154.  A  ROOT  of  an  equation  is  such  a  value  of  the  un- 
known quantity  as,  being  substituted  for  it,  will  satisfy  the 
equation  ;  that  is,  make  the  two  members  equal. 

The  SOLUTION  of  an  equation  is  the  operation  of  finding 
its  roots. 

INCOMPLETE    EQUATIONS. 

155.  It  may  happen,  that  2/>,  the  coefficient  of  the  first 
power  of  x,  in  the  equation  x  +  2px  =  q,   is  equal  to  0. 
In  this  case,  the  first  power  of  x  will  disappear,  and  the 
equation  will  take  the  form, 

*=! (1.) 

This  is  called  an  incomplete  equation ;  hence, 

AN  INCOMPLETE  EQUATION,  when  reduced,  contains  but 
two  terms;  the  square  of  the  unknown  quantity,  and  a 
known  term. 

156.  Extracting  the  square  root  of  both  members  of 
Equation  ( 1 ),  we  have, 

x  =    ±i/q. 

154.  What  is  the  root  of  an  equation  ?    What  is  the  solution  of  an 
equation  ? 

155.  What  form  will  the  reduced  equation  take  when  the  coefficient  ol 
a:  is  0  ?     What  is  the  equation  then  called  ?     How  many  terms  are  there 
iu  an  incomplete  equation  ?     What  are  they  ? 

156.  What  is  the  rule  for  the  solution  of  an  incomplete  equation  T 
How  many  roots  are  there  in  every  incomplete  equation  ?     How  do  tb« 
roots  compare  with  each  other  ? 


EQUATIONS     OF     THE     SECOND     DEGREE.      201 

Hence,  for  the  solution  of  incomplete  equations : 

RULE. 

I.   Reduce  the  equation  to  the  form  x2  =  q: 
II.    Then  extract  the  square  root  of  both  members. 

NOTE. — There  will  be  two  roots,  numerically  equal,  but 
having  contrary  signs.  Denoting  the  first  by  a;',  and  the 
second  by  «",  we  have, 

x'  —   +i/q,     and    x"  =  —  *<fq. 

VERIFICATION. 

Substituting  +  -y/^,  or  —  i/q,  for  JB,  in  Equation  ( 1 ), 
we  have, 

(+V5)2  =  q;     and,     (-V?)2  =  <?; 

hence,  both  satisfy  the  equation ;  they  are,  therefore,  roots. 
(Art.  154.) 

EXAMPLES. 

1.  What  are  the  values  of  x  in  the  equation, 

Sx2  -f  8  =  5x2  —  10  ? 
By  transposing,  3x2  —  5x2  =  —  10  —  8. 
Reducing,  —  2x2  =  —  18. 

Dividing  by  —  2,  »2  =  9. 

Extracting  square  root,     x  =   ±  -y/9  =   +  3   and   —  8. 
Hence,  x'  =   +3,   and  x"  =   —  3. 

2.  What  are  the  roots  of  the  equation, 

3«2+  6  =  4a2-  10? 

Ans.  x'  =   +4,    x"  =  —  4, 


202  ELEMENTARF      ALGEBKA. 

8.  What  are  the  roots  of  the  equation, 


Ans.   x'  —   +  9,    x"  =   -  9. 

I.  What  are  the  roots  of  the  equation, 
4<e2  +  13  —  2x2  =  45  ? 

Ans.   x'  =   +4,    x"  =   —  4. 

5.  What  are  the  roots  of  the  equation, 

Gx"  —  7  =  3x2  +  5  ? 

Ans.  .  x'  =   +  2,    x"  =    -  2, 

6.  What  are  the  roots  of  the  equation, 

x2 

8  4-  5x2  =  -  +  4x2  -|-  28  ? 
5 

Ans.   x'  —   +5,    *"  =   —  5. 

7.  What  are  the  roots  of  the  equation, 

^+5   _  *2+29    =   m_5a.2? 

Ans.  .x'=   +5,    x"  =   —  6. 

8.  What  are  the  roots  of  the  equation, 

jB2  +  ab  =  So2  ? 


.  a    = 
9.  What  are  the  roots  of  the  equation, 

a;2  =  5  -f  a^  ? 


'«  -  25  Va  -  2* 


PROBLEMS. 


PROBLEMS. 

1.  What  number  is  that  which  being  multiplied  by  itself 
the  product  will  be  144  ? 

Let  x  =  the  number :  then, 

x  X  x  =  xz  =  144. 

It  is  plain  that  the  value  of  x  will  be  found  by  extracting 
the  square  root  of  both  members  of  the  equation :  that  is, 

T/x*  =   yT*4:  that  is,    x  =  12. 

2.  A  person  being  asked  how  much  money  he  had,  said, 
if  the  number  of  dollars  be  squared  and  6  be  added,  the  sum 
will  be  42  :  how  much  had  he  ? 

Let  x  =    the  number  of  dollars. 

Then,  by  the  conditions, 

xz  +  6   =  42 ; 

hence,  x2  =  42  —  6  =  36, 

and,  x  =     6.  Ans.   $6. 

3.  A  grocer  being  asked  how  much  sugar  he  had  sold  to 
a  person,  answered,  if  the  square  of  the  number  of  pounds 
be  multiplied  by  7,  the  product  will  be  1575.    How  many 
pounds  had  he  sold  ? 

Denote  the  number  of  pounds  by  x.     Then,  by  the  con- 
ditions of  the  question, 

7«3  ='  1575 ; 

hence,  x2  =     225, 

and,  x   =       15.  Ans.    15. 

4.  A  person  being  asked  his  age,  said,  if  from  the  square 


204  ELEMENTARY        ALGEBKA. 

of  my  age  in  years,  you  take  192  years,  the  remainder 
be  the  square  of  half  my  age :  what  was  his  age  ? 

Denote  the  number  of  years  in  his  age  by  x. 

Then,  by  the  conditions  of  the  question, 

/I    \          x2 
y?  —  192   =   (-a;]2  =  — , 

and  by  clearing  the  fractions, 

402  —  768  —  x2 ; 

hence,  4ce2  —  cc2  =   768, 

and,  3JC2  =  768, 

a?  =  256, 
x   —     16.         Ans.   16  years. 

5.  What  number  is  that  whose  eighth  part  multiplied  by 
its  fifth  part  and  the  product  divided  by  4,  will  give  a  quo- 
tient equal  to  40  ? 

Let  x  =  the  number. 

By  the  conditions  of  the  question, 


hence,  =  40  ; 

by  clearing  of  fractions, 

x2  =  6400, 

x   =       80.  Ans.    80. 

6.  Find  a  number  such  that  one-third  of  it  multiplied  by 
one  fourth  shall  be  equal  to  108.  Ans.   36. 

7.  What  number  is  that  whose  sixth  part  multiplied  by 
its  fifth  part  and  the  product  divided  by  ten,  will  give  a 
quotient  equal  to  3  ?  Ans.   30. 


PROBLEMS.  205 

8.  "What  number  is  that  whose  square,  plus  18,  -will  be 
equal  to  half  the  square,  plus  30£  ?  Ans.   5. 

9.  What  numbers  are  those  which  are  to  each  other  as 
1  to  2,  and  the  difference  of  whose  squares  is  equal  to  75  ? 

Let     x  =  the  less  number. 
Then,    2x  =  the  greater. 

Then,  by  the  conditions  of  the  question, 

4a52  -  y?  =  75  ; 

hence,  3«2  =  75, 

and  by  dividing  by  3,    y?  =  25,    and    x  =  5, 

and,  2«  =  10. 

Ans.   5  and  10. 

10.  What  two  numbers  are  those  which  are  to  each  other 
as  5  to  6,  and  the  difference  of  whose  squares  is  44  ? 

Let     x  =  the  greater  number. 

Then,   -x  =  the  less. 
6 

By  the  conditions  of  the  problem, 

3.2 3.2 A  A  . 

36 

by  clearing  of  fractions, 

36a;2  —  25ar»  =  1584 ; 

hence,                                   lla;2  =  1584, 

and,                                          aJ2^  144 ; 

hence,                                        x    =  12, 

and,  -x   =       10. 

6 

Ans.   10  and  12. 


206  ELEMENTARY       ALGEBRA. 

11.  What  two  numbers  are  those  which  are  to  each  other 
as  3  to  4,  and  the  difference  of  whose  squares  is  28  ? 

Ans.    6  and  8. 

12.  What  two  numbers  are  those  which  are  to  each  other 
as  5  to  11,  and  the  sum  of  whose  squares  is  584  ? 

Ans.    10  and  22-. 

13.  A  says  to  .Z?,  my  son's  age  is  one  quarter  of  yours, 
and  the   difference  between  the   squares  of  the   numbers 
representing  their  ages  is  240  :  what  were  their  ages  ? 

I  Eldest,       16, 
Ans.     \  „ 

(  lounger,     4. 

Tico  unJcnoicn  quantities. 
15 7.     When  there  are  two  or  more  unknown  quantities: 

I.  Eliminate   one  of  the  unknown  quantities  by  Art. 
113: 

IL  Then  extract  the  square  root  of  both  members  of  the 
equation. 

PEOBLEMS. 

1.  There  is  a  room  of  such  dimensions,  that  the  difference 
of  the  sides  multiplied  by  the  less,  is  equal  to  36,  and  the 

product  of  the  sides  is  equal  to  360  :  what  are  the  sides? 

» 
Let  x  —   the  length  of  the  less  side ; 

y  =   the  length  of  the  greater. 
Then,  by  the  first  condition, 

(y  —  x)x  =     36  ; 
and  by  the  2d,  xy  =  360. 

157.  How  do  you  proceed  when  there  are  two  or  more  unknown  quan- 
tities ? 


PROBLEMS.  207 

From  the  first  equation,  we  have, 

xy  —  xz  =     36  ; 

and  by  subtraction,  x2  =  324. 

Hence,  x  =   -y/324  =     18; 

360 

y  =  -   -  =     20. 
18 

Ans.   a;  =   18,    y  —   20. 

2.  A  merchant  sells  two  pieces  of  muslin,  which  together 
measure  12  yards.     He  received  for  each  piece  just  so  many 
dollars  per  yard  as  the  piece  contained  yards.     Now,  he  gets 
four  times  as  much  for  one  piece  as  for  the  other :  how  many 
yards  in  each  piece  ? 

Let    x  =    the  number  of  yards  in  the  larger  piece ; 
y  =    the  number  of  yards  in  the  shorter  piece. 
Then,  by  the  conditions  of  the  question, 

x  +  y  =  12. 

x  X  x  =  x2  =  what  he  got  for  the  larger  piece ; 
y  x  y  =  y2-  =  what  he  got  for  the  shorter ; 
and,  as2  =  4y2,  by  the  2d  condition, 

x   =  2y,    by  extracting  the  square  root. 
Substituting  this  value  of  x  in  the  first  equation,  we  have, 

y  +  2y  =  12  ; 
and,  consequently,  y  =  4, 

and,  x  =  8. 

Ans.  8  and  4. 

3.  What  two  numbers  are  those  whose  product  is  30,  and 
the  quotient  of  the  greater  by  the  less,  3i  ?    Ans.  10  and  3. 

4.  The  product  of  two  numbers  is  a,  and  their  quotient 

b :  what  are  the  numbers  ?  

Ans.  yob,   and 


208  ELEMENTAET      ALGEBKA. 

5.  The  sum  of  the  squares  of  two  numbers  is  117,  and  the 
difference  of  their  squares  45  :  what  are  the  numbers? 

Ans.  9  and  6. 

6.  The  sum  of  the  squares  of  two  numbers  is  a,  and  the 
difference  of  their  squares  is  b  :  what  are  the  numbers  ? 

A                    fa  +  b                 fa  —  b 
Ans.   x  =  v  / ,    y  =  \  / . 

V       2  V       2 

7.  What  two  numbers  are  those  which  are  to  each  othei 
as  3  to  4,  and  the  sum  of  whose  squares  is  225  ? 

Ans.  9  and  ^2. 

8.  What  two  numbers  are  those  which  are  to  each  other 
as  m  to  n,  and  the  sum  of  whose  squares  is  equal  to  a2  ? 

ma  na 

Ans. 


9.  What  two  numbers  are  those  which  are  to  each  other 
as  1  to  2,  and  the  difference  of  whose  squares  is  75  ? 

Ans.  5  and  10. 

10.  What  two  numbers  are  those  which  a*re  to  each  other 
as  m  to  w,  and  the  difference  of  whose  squares  is  equal  to  b2  ? 

mb  nb 

Ans.  — 


11.  A  certain  sum  of  money  is  placed  at  interest  for  six 
months,  at  8  per  cent,  per  annum.    Now,  if  the  sum  put  at 
interest  be  multiplied  by  the  number  expressing  the  interest, 
the  product  will  be  $562500 :  what  is  the  principal  at  in- 
terest? Ans.  $3750. 

12.  A  person  distributes  a  sum  of  money  between  a  num- 
ber of  women  and  boys.     The  number  of  women  is  to  the 
number  of  boys  as  3  to  4.    Now,  the  boys  receive  one-half 
as  many  dollars  as  there  are  persons,  and  the  women,  twice 
as  many  dollars  as  there  are  boys,  and  together  they  receive 


COMPLETE      EQUATIONS.  209 


J38  dollars:  how  many  women  were  there,  and  how  many 
boys? 

36  women. 

48  boys. 

COMPLETE    EQUATIONS. 


Ans.  j 


158.  The  reduced  form  of  the  complete  equation  (Art. 
153)  is, 

a2  +  2px  =  q. 

Comparing  the  first  member  of  this  equation  with  the 
square  of  a  binomial  (Art.  54),  we  see  that  it  needs  but  the 
square  of  half  the  coefficient  of  #,  to  make  it  a  perfect  square. 
Adding  p2  to  both  members  (Ax.  1,  Art.  102),  we  have, 

y?  +  2px  -f-  P"  =  q  +  pz. 

Then,  extracting  the  square  root  of  both  members  (Ax.  5), 
we  have, 


x  +  p  =    ±  iq  +  p2. 
Transposing  p  to  the  second  member,  we  have, 

x  =   —  p  ±  -\/q  +  pz. 

Hence,  there  are  two  roots,  one  corresponding  to  the  plus 
sign  of  the  radical,  and  the  other  to  the  minus  sign.  De- 
noting these  roots  by  x'  and  x",  we  have, 

x'  —   —  p  +  ^q  +  p2,    and    x"  =   —  p  —  -\/q  -f  p2. 

The  root  denoted  by  x'  is  called  the  first  root  ;  that  de- 
noted by  x"  is  called  the  second  root. 

1  58.  What  is  the  form  of  the  reduced  equation  of  the  second  degree  ? 
What  is  the  square  of  the  binomial  x  +  p  ?  How  many  of  those  terms 
are  found  in  the  first  term  of  the  reduced  equation?  What  must  be 
added  to  make  the  first  member  a  perfect  square  ?  How  many  roots  are 
there  in  every  equation  of  the  first  degree?  What  is  the  first  root  equal 
to  ?  What  is  the  second  equal  to  ? 


210  ELEMENTARY       ALGEBUA 

159.  The  operation  of  squaring  half  the  coefficient  of 
x  and  adding  the  result  to  both  members  of  the  equation,  is 
called  Completing  the  Square.  For  the  solution  of  every 
complete  equation  of  the  second  degree,  we  have  the  fol- 
lowing 

KULE. 
I.   Reduce  the  equation  to  the  form,  xz  +  2px  =  q: 

IE.  Take  half  the  coefficient  of  the  second  term,  square 
it,  and  add  the  result  to  both  members  of  the  equation : 

HI.  Then  extract  the  square  root  of  both  members  ;  after 
ivhich,  transpose  the  known  term  to  the  second  member. 

NOTE. — Although,  in  the  beginning,  the  student  should 
complete  the  square  and  then  extract  the  square  root,  yet 
he  should  be  able,  in  all  cases,  to  write  the  roots  immediately, 
by  the  following  (See  Art.  158) 

ETTLE. 

I.  The  first  root  is  equal  to  half  the  coefficient  of  the 
second  term  of  the  reduced  equation,  taken  with  a  contrary 
sign,  plus  the  square  root  of  the  second  member  increased 
by  the  square  of  half  the  coefficient  of  the  second  term  : 

II.  The  second  root  is  equal  to  half  the  coefficient  of  the 
second  term  of  the  reduced  equation,  taken  with  a  contrary 
sign,  minus  the  square  root  of  the  second  member  increased 
by  the  square  of  half  the  coefficient  of  the  second  term. 

160.     We  will  now  show  that  the  complete  equation  of 

159.  What  is  the  operation  of  completing  the  square?     How  many 
operations  are  there  in  the  solution  of  every  equation  of  the  second  de- 
gree ?     What  is  the  first ?     What  the  second?     What  the  third  ?     Give 
the  rule  for  writing  the  roots  without  completing  the  square? 

160.  How  many  forms  will  the  complete  equation  of  the  second  degree 
assume?     On  what  will  these  forms  depend?     What  are  the  signs  of  2p 


COMPLE1E      EQUATIONS.  211 

the  second  degree  will  take  four  forms,  dependent  on  the 
signs  of  2p  and  q. 

1st.  Let  us  suppose  2p  to  be  positive,  and  q  positive;  we 
shall  then  have, 

a2  +  2px  =  q.        .    .     .     .     (l.) 

2d.  Let  us  suppose  2p  to  be  negative,  and  q  positive ; 
we  shall  then  have, 

x7  —  2px  =  q (2.) 

3d.  Let  us  suppose  2p  to  be  positive,  and  q  negative ; 
\ve  shall  then  have, 

«2  +  2px  =    —  q.         .     .     .     ( 3.) 

4th.  Let  us  suppose  2p  to  be  negative,  and  q  negative ; 
we  shall  then  have, 

x2  —  2px  =   —  q.        ...     (4.) 

As  these  are  ah1  the  combinations  of  signs  that  can  take 
plane  between  2p  and  q,  we  conclude  that  every  complete 
equation  of  the  second  degree  will  be  reduced  to  one  or  the 
other  of  these  four  forms : 

jc2  +  2px  =  +  <7,  .  .  1st  form. 

x*  —  2px  =  +  §-,  .  .  2d   form. 

xz  -f  2pa;  —  —  y,  '  .  .  3d   form. 

a2  —  2px  =  —  q,  ,  .  4th  form. 

EXAMPLES    OF    THE    FIRST    FOEM. 

1.  What  are  the  values  of  x  in  the  equation, 

2x>  +  8x  =  64  ? 

If  we  first  divide  by  the  coefficient  2,  we  obtain 
«2  -f-  4x  =  32. 

and  q  in  the  first  form?    What  in  the  second?    What  in  the  third? 
WViat  in  the  fourth  ? 


212  ELEMENTARY      ALGEBKA. 

Then,  completing  the  square, 

x2  +  4x  +  4  =  32  +  4  =  36. 
Extracting  the  root, 

x  +  2  =   ±  -v/36  =  +  6,    and    —  6. 

Hence,  x'  =   —  2  +  6  =  +4; 

and,  x"  —   —  2  —  6  =  —  8. 

Hence,  in  this  form,  the  smaller  root,  numerically,  is  positive, 
and  the  larger  negative. 

VERIFICATION. 

If  we  take  the  positive  value,  viz. :  xf  =   +  4, 
the  equation,  a;2  +  4&  =  32, 

gives  42  +  4  x  4  =  32  ; 

and  if  we  take  the  negative  value  of  »,  viz.  :«,"=.—  8, 
the  equation,  a2  +  4x  =  32, 

gives  (—  8)2  +  4(—  8)   =  64  -  32   =  32  ; 

from  which  we  see  that  either  of  the  values  of  x,  viz.: 
x'  =   +  4,   or  x"  —   —  8,    will  satisfy  the  equation. 

2.  What  are  the  values  of  x  in  the  equation, 

3«2  +  12aj  —  19  =   —  a2  —  12»  +  89? 
By  transposing  the  terms,  we  have, 

3a2  +  y?  +  12cc  +  I2x  =  89  +  19 ; 

and  by  reducing, 

4z2  +  24o;  =  108; 

and  dividing  by  the  coefficient  of  jc2, 

a;2  +  Qx  =  27. 


C  O  M  P  L  K  T  K      EQUATIONS.  213 

Now,  by  completing  the  square, 

x2  +  6x  +  9  =  36 ; 
extracting  the  square  root, 

x  +  3  =   ±-/36  =   +  6,   and    -•  6; 
hence,  xr  =+6  —  3   =   -f  3; 

and,  x"  —   —  6  —  3   =  —  9. 

VERIFICATION. 

If  we  take  the  plus  root,  the  equation, 

cc2  -f  6z  =  27, 

gives  (3)2+  6(3)  =  27; 

and  for  the  negative  root, 

xz  +  Qx  =  27, 
gives  (-  9)2  +  6(—  9)   =   81  —  54  =  27. 

3.  What  are  the  values  of  a;  in  the  equation, 

xz  -  lOaj  +  15  =  ^  -  34a;  +  155  ? 

O 

By  cleaiing  of  fractions,  we  have, 

5xz  —  50x  +75  =  xz  —  I70a  +  775; 
by  transposing  and  reducing,  we  obtain, 
4xz  +  12Cte  =  700 ; 
then,  dividing  by  the  coefficient  of  «2,  we  have, 

xz+    30x  =  175; 
and  bj  completuig  the  square, 

y?  +  30a  -f-  225   =  400 ; 


214  ELEMENTARY      ALOEBKA. 

and  by  extracting  the  square  root, 

x  +  15   =   ±-y/40()   =    +  20,    and    —  20. 
Hence,  xf  =   +  5,   and  x"  =   —  35. 

VERIFICATION. 

For  the  plus  value  of  #,  the  equation, 
x2  +  30x  =  175, 
gives,  (5)2  -f  30  X  5   =  25  +  150  =  175. 

And  for  the  negative  value  of  £C,  we  have, 

(_  35)2  _j_  30(_  35)   _   1225  —  1050   —   175. 

4.  What  are  the  values  of  x  in  the  equation, 

s*-3»+!  =  '-!— *+T?' 

Clearing  of  fractions,  we  have, 

10a2  -  6x  +  9   =   96  —  Sx  —  I2xz  +  273; 
transposing  and  reducing, 

22a2  +  2x  =  360 ; 
dividing  both  members  by  22, 

2  360 

£C2  H X    =    • 

r  22  22 

(1  \2 
—  I    to  both  members,  and  the  equation  becomes, 

2          /  1  \2        360       /  1  \2 

/>•*    _l       _,__/>•    J,  |        __      _  I    • 

^22*    h  \22/       '    22     hl22/' 
whence,  by  extracting  the  square  root, 

-         J_  /360       /  1  y 

h  22   Z        :  V  22    "h  \22/' 


COMPLETE      EQUATIONS.  215 

therefore, 


/  1  \2 
\22/' 


and,  «"  =   __ 

It  remains  to  perform  the  numerical  operations.     In  the 
first  place, 

860/1  \2 
22    "h  \22/' 

must  be  reduced  to  a  single  number,  having  (22)2  for  its 
denominator.     Now, 


360        /J_\2  _ 
"22"        \22/    : 


360  X  22  +  1          7921 


22         \22/  (22)2  "   (22)2' 

extracting  the  square  root  of  7921,  we  find  it  to  be  89; 
therefore, 

89 


22         \22 

Consequently,  the  plus  value  of  x  is, 


-1     L   —  — 

o2  +  22   ~   22   ~~     ' 


and  the  negative  value  is, 


1     _  89  45 

22  ~  22  ~   ~  TI 


that  is,  one  of  the  two  values  of  x  which  will  satisfy  the 
proposed  equation  is  a  positive  whole  number,  and  the  other 
a  negative  fraction. 

NOTE. — Let  the  pupil  be  exercised  in  writing  the  roots,  in 
Jie  last  five,  and  in  the  following  examples,  without  com- 
pleting the  square. 


216  ELEMENTARY      ALQEBKA. 

6.  What  are  the  values  of  x  in  the  equation, 

SaJ2  +  2x  —  9  =  76  ? 

x'  =  5. 


6.  What  are  the  values  of  a;  in  the  equation, 

K/v.  /vt2 

2x2  +  8a;  +  7  =  —  -  ^  +  197  ? 
4          8 


' 


=   8. 


7.  What  are  the  values  of  x  in  the  equation, 


=:    9. 

'    (  x"—   —  64}. 
8.  What  are  the  values  of  a;  in  the  equation, 

x2       5x  x 

T        4~  "  2  ~  '    *' 

'   =    2. 


9.  What  are  the  values  of  a;  in  the  equation, 
9?   .   x        xz        x        13 


EXAMPLES    OF  THE   SECOND   FORM. 

1.  What  are  the  values  of  x  in  the  equation, 
a-2  —  Sx  -f-  10  =  19? 


COMPLETE      EQUATIONS.  217 

By  transposing, 

«2  -  Sx  =  19  —  10  =     9; 
then,  by  completing  the  square, 

a;2  -  8x  +  16  =     9  +  16   =  25  ; 
and  by  extracting  the  root, 

x  —  4   =    ±  i/25  =   +  5,    or    —  5. 
Hence, 

x'  =  4  +  5   —  9,    and    x"  =  4  —  5  =.  •   —  1. 

That  is,  in  this  form,  the  larger  root,  numerically,  u 
positive,  and  the  lesser  negative. 

VERIFICATION. 

If  we  take  the  positive  value  of  x,  the  equation, 
a2  —  8x  =  9,    gives     (9)2  —  8x9  =  81  —  72   =  9; 
and  if  we  take  the  negative  value,  the  equation, 
jc2  -  8a;  =  9,    gives     (-  I)2  —  8  (—  1)  =  1  +  8  =  9; 

from  which  we  see  that  both  roots  alike  satisfy  the  equa- 
tion. 

2.  What  are  the  values  of  x  in  the  equation, 


By  cleaiing  of  fractions,  we  have, 

6*2  +  4x  —  180  =  3xz  +  12*  —  177  , 
and  by  transposing  and  reducing, 

3x2  —  8x  =  3  ; 
and  dividing  by  the  coefficient  of  a;2,  we  obtain, 

*•-!*  =  '• 


218  ELEMENTARY       ALGEBRA. 

Then,  by  completing  the  square,  we  have, 

^LV+21      h^--- 

Vit  **J        ^  —  .1  1  —  • 

39  99 

and  by  extracting  the  square  root, 

4  /25  5 


Hence, 

=  3  +  3  =          3) "3       3  ~        S 

VEKIFICATION. 

For  the  positive  root  of  a;,  the  equation, 


Q 

gives  32  --  x3  =  9  —  8  =  1; 

3 

and  for  the  negative  root,  the  equation, 
xz-  ~x  =  1, 

1\2      8  1         1,8 

3J  -  3  X  ~  3   =  9  +  9  =  l 

3.  What  are  the  values  of  x  in  the  equation, 


Clearing  of  fractions,  and  dividing  by  the  coefficient  of 
x2,  we  have, 

*•-•  =  n- 


COMPLETE      EQUATIONS.  219 

Completing  the  square,  we  have, 

2  -  -    +  -  I   "     1?. 

3          9   ~  9   ~~   36  ' 

then,  by  extracting  the  square  root,  we  have, 

1  /49  7  7 

3~~        V  36   ~          6'  6  J 

hence, 

,_17         9  „  _  \        7  5 

O  O  O  O  O  O 

VERIFICATION. 

If  we  take  the  positive  root  of  JB,  the  equation, 

x        ^jc  =.  14, 

2 
gives  (H)2 X  1£  =  2J  —  1  =  Ij.; 

O 

and  for  the  negative  root,  the  equation, 


52      2  5         25        10         45 


4.  What  are  the  values  of  a;  in  the  equation, 

4a2  —  2a2  +  2craj  =  I8at>  — 


By  transposing,  changing  the  signs,  and  dividing  tfj-  2, 
the  equation  becomes, 

a2  —  ax  =  2a2  —  9a&  +  952  ; 


220  ELEMENTARY      ALGEBRA. 

whence,  completing  the  square, 


-  **  +  7  =  -T  ~ 


extracting  the  square  root, 


/y»     — .    

VJ         


Now,  the  square  root  of  —  —  Qab  +  95'^,    is  evidently 

i 

35.     Therefore, 

s'  =       2a  —  3b. 
fj'--  —    a  +  3b. 

What  will  be  the  numerical  values  of  cc,  if  we  suppose 
a  =  6,   and    5  =  1? 

5.  What  are  the  values  of  a:  in  the  equation, 


4 

xz  —   45  — 

5 


within 


(x'    =        7.12  Ho  withi 
AnS'\x»=    -5.73f     0.01. 


6.  What  are  the  values  of  x  in  the  equation, 

8xz  —  Ux  +10  =  2<e  +  34? 


7.  What  are  the  values  of  x  in  the  equation, 

c'    =   8. 


-  30  +  x  =  2x  -  22  ? 
4 


A 
Ans. 

x"  =    —  4. 


8.  What  are  the  values  of  x  in  the  equation, 


C  O  M  1'  L  E  T  E       EQUATIONS. 

9.  What  are  the  values  of  x  in  the  equation, 
2ax  —  x2  =    —  2ab  —  bz? 

j  x'    =  '2a  +  ft. 


10.  "What  are  the  values  of  x  in  the  equation, 


EXAMPLES    OF    THE    THIKD    FOKM. 

1.  What  are  the  values  of  a;  in  the  equation, 

x2  +  4x  —   —  3  ? 
First,  by  completing  the  square,  we  have, 

£C2  +  4x  +  4   =    —  3  +  4   =   1; 
and  by  extracting  the  square  root, 

as+2=±-v/I   =   +l>   and    —  1 ; 
hence,  x'  =  — 2  +  1  =  —  1;   and  x"  =  —  2  —  1  =  —  8. 

That  is,  in  this  form  both  the  roots  are  negative. 

VERIFICATION. 

If  we  take  the  first  negative  value,  the  equation, 

x2  +  4x  =   —  3, 

gives  (—  I)2  +  4(—  1)  =  1—  4=—  3; 

and  by  taking  the  second  value,  the  equation, 


222  E  L  E  M  K  N  T  A  R  Y       ALGEBRA. 

gives  (—  3)2  +4(  —  3)   -9  —  12  =   —  3; 

hence,  both  values  of  x  satisfy  the  given  equation. 

2.  What  are  the  values  of  a;  in  the  equation, 

—  Y  —  5x  —  16   =  12  +  -z2  +  6z? 

By  transposing  and  reducing,  we  have, 
—  x2  —  lice  i=  28; 
then,  dividing  by  —  1,  the  coefficient  of  £2,  we  have, 

xz  +  Hxi=   —  28; 
then,  by  completing  the  square, 

x*  +  llx  +  30.25   =   2.25; 


hence,       x  +  5.5   =   ±-y/2.25  =   +  1.5,   and    —  1.5; 
consequently,       x'  =   —  4,   and  x"  =   —  7. 

3.  What  are  the  values  of  a;  in  the  equation, 

-  |2  -  2x  -  5  =  Ix*  +  5a;  +  5  ? 

x'    =   —  2. 

4.  What  are  the  values  of  a;  in  the  equation, 

O/J.2       I         Oy,       02.     _    .     -Cf  9 

ZK     -f-    SX     —  L%  A  I 


.. 

as"  =   —  4. 
5.  What  are  the  values  of  x  hi  the  equation, 


4a;2  +  \x  +  3a;  =   -  14»  -  3}  - 
5 


COMPLETE      EQUATIONS.  223 

6.  What  are  the  values  of  a:  in  the  equation, 

—  y?  —  4  -  -x  =  ^  +  24a;  +  2  ? 

(  x'   =  —  J- 
Ans.  \ 

(x"  =   —  8. 

7.  What  are  the  values  of  a;  in  the  equation, 

ic2  +  to  +  20  =   -    |e2  —  lla;  -  60  ? 

'  (  x"  =   -  10. 

8.  What  are  the  values  of  x  in  the  equation, 


9.  What  are  the  values  of  a;  in  the  equation, 

4  T  1  3 

,^/y»2      I        K  rt/t       I  ^—  ^_/>»2    CI  JL'T*    - 

»O         [^     t7*O    ^^  ^^  f*^      """""          1  "0" 

5  45  4 

Ans. 

10.  What  are  the  values  of  x  in  the  equation, 

x  -  y?  —  3  —  Qx  +  1  ? 

j  x'    =   —  1. 
AnS'  \  x"  =   -  4! 

11.  What  are  the  values  of  *  in  the  equation, 

x*  +  4a;  -  90  =   —  93  ? 


EXAMPLES    OF    THE    FOURTH    FOKM. 

1.  What  are  the  values  of  x  in  the  equation, 
x2  —  8a;  =  —  7  ? 


224:  ELEMENTARY      ALGEBRA. 

By  completing  the  square,' we  have, 

x2  -  Sx  +  16  =    —  7  +  16  =  0; 
then,  by  extracting  the  square  root, 

x  —  4  =   ±i/9  =   +  3,    and    —  3 ; 
hence,  x'  =   +  7,   and  x"  =   +  !• 

That  is,  in  this  form,  both  the  roots  are  positive. 

VERIFICATION. 

If  we  take  the  greater  root,  the  equation, 
a;2  —  8a;  =   —  7,     gives,    72  —  8x7   =  49  —  56  :  z  —  7 ; 
and  for  the  lesser,  the  equation, 

x*  -  Sx  -     -  7,    gives,    I2 -8x1  =  1  —  8=  -7; 
hence,  both  of  the  roots  wih1  satisfy  the  equation. 

2.  What  are  the  values  of  x  in  the  equation, 

40 

—  lia;2  +  3x  -  10  =  Ha2  -  IBx  +  —  ? 

2 

By  clearing  of  fractions,  we  have, 

—  So;2  +  Qx  —  20  —  Sx2  -  36z  +  40 ; 
then,  by  collecting  the  similar  terms, 

_  6cc2  +  42a;  =  60  ; 

then,  by  dividing  by  the  coefficient  of  cc2,   which  is  —  0, 
we  have, 

a:2  -  Vaj  =    -  10. 

By  completing  the  square,  we  have, 

x7'  —  7z  +  12.25   =-   2.25, 


COMPLETE      EQUATIONS.  225 

and  by  extracting  the  square  root  of  both  members, 

x  —  3.5   =   ±y^25  =   +  1.5,   and    —  1.5; 
hence, 

x'  =  3.5  +  1.5   =  5,    and     x"  =  3.5  -  1.5   =  2. 

VERIFICATION. 

If  we  take  the  greater  root,  the  equation. 
xz  —  fce  =  -  10,    gives,    52  -  7  X  5  =  25  —  35  •—    -  10; 
and  if  we  take  the  lesser  root,  the  equation, 
x2  —  7x  =  —  10,     gives,     22  —  7  X  2  =  4  —'14  =•    -  10. 

3.  What  are  the  values  of  x  in  the  equation, 

-  Bx  +  2a2  +  1   =   17fe  -  2z2  -  3  ? 
By  transposing  and  collecting  the  terms,  we  have, 

4«2  —  20|a;  =   —  4  ; 
then  dividing  by  the  coefficient  of  *2,  we  have, 

£C2  -   5}X    =     -    I. 

By  completing  the  square,  we  obtain, 

,    169  169         144 

»-i4«,+  _==  -i  +  -5-  =  — ; 

and  by  extracting  the  root, 


hence, 

12  12          1 

*  =  2f  -f  —   =  5,    and,    x"  =  2J  -  --  =  -. 

VERIFICATION. 

If  we  take  the  greater  root,  the  equation, 
a;2  —  5$x  =  —  1,    gives,   52  —  5£  x  5  =  25  —  26  =  — • 


226  ELEMENTARY      ALGEBRA. 

and  if  we  take  the  lesser  root,  the  equation, 
*  -5lx=-  1,  gives,  Q)2-  6jxJ=^-!  =  - 
4.  What  are  the  values  of  x  in  the  equation, 


fa'    =  3. 


5.  What  are  the  values  of  a  in  the  equation, 

—  4a2  —    x  +  1     =   —  5a2  +  8a? 


6.  What  are  the  values  of  a  in  the  equation, 


7.  What  are  the  values  of  a  in  the  equation, 

a2  —  10Tyc  =   —  1  ? 

(a'    = 

AllS-   \  vn   _ 
I  a     = 

8.  What  are  ;he  values  of  a  in  the  equation, 


=  10. 


2a2 

—  27a  ^   —  —  +  100  =   -—  +  12a  —  26  ? 
5  5 


f  =     ' 

Ans.  i    „ 

(  x"  =  6. 


9.  What  are  the  values  of  a  in  the  equation, 

—  —  22a  -f  15   =    —  — -  -f  28a  —  30  ? 
8  3 


x'    =   9. 


PROPERTIES     OF     EQUATIONS.  227 

10.  What  are  the  values  of  x  in  the  equation, 

2»2  -30a;-|-3  =   -  »2  -f-  3-frx  —  ^-  ? 

f   —  11 

fc      —       J.  A» 


PROPERTIES    OF  EQUATIONS  OF  THE   SECOlfD    DEGREE. 
FIRST  PROPERTY. 

161.    We  have  seen  (Art.  153),  that  every  complete 
equation  of  the  second  degree  may  be  reduced  to  the  form, 

«2  +  2px  =  q  .....     (  1.) 

Completing  the  square,  we  have, 

x2  +  2px  +  p2  —  q  -f  pz  ; 
transposing  q  -f  p2  to  the  first  member, 

tc2  +  2px  +  p*  —  (q  +  j(?2)   =  0.        .     (  2.) 
Now,  since  as2  +  2px  +  ^>2  is  the  square  of  x  +  ^?,  and 


q  +  j92  the  square  of  ^/g~+  pz,  we  may  regard  the  first 
member  as  the  difference  between  two  squares.  Factoring, 
(Art.  56),  we  have, 

(x+p  +  V  3  +  Pz)  (x+p—  V<1  +  P*)  =  0.    .     (  3.) 

This  equation  can  be  satisfied  only  in  two  ways  : 

1st.  By  attributing  such  a  value  to  x  as  shall  render  the 
first  factor  equal  to  0  ;  or, 

161.  To  what  form  may  every  equation  of  the  second  degree  be  re- 
duced? What  form  will  this  equation  take  after  completing  the  square 
and  transposing  to  the  first  member  ?  After  factoring  ?  In  how  many 
wajc  may  Equation  (  3  )  be  satisfied  ?  What  are  they  ?  How  many  roots 
has  every  equation  of  the  s^ond  degree? 


228  ELEMENTARY      ALGEBRA. 

2d.  By  attributing  such  a  value  to  x  as  shall  raider  ih« 
second  factor  equal  to  0. 

Placing  the  second  factor  equal  to  0,  we  have, 
x+p  —  -\/q+pz  =  0;  and  x'  =  —  p+  \/Q  +P2-        (4-) 
Placing  the  first  factor  equal  to  0,  we  have, 

"  =  —p  — 


Since  every  supposition  that  will  satisfy  Equation  (  3  ), 
also  satisfy  Equation  (  1  ),  from  which  it  was  derived,  it  fol- 
lows, that  x'  and  x"  are  roots  of  Equation  (  1  )  ;  also,  that 

Every  equation  of  the  second  degree  has  two  roots,  and 
only  two. 

NOTE.  —  The  two  roots  denoted  by  x'  and  »",  are  the 
same  as  found  in  Art.  158. 

SECOND    PROPERTY. 

• 

162.  We  have  seen  (Art.  161),  that  every  equation  ol 
the  second  degree  may  be  placed  under  the  form, 

(x  +  p  +  Vg  +  PZ)  (x  +  P  —  V<i  +P2)  =  °- 

By  examining  this  equation,  we  see  that  the  first  factor 
may  be  obtained  by  subtracting  the  second  root  from  the 
unknown  quantity  x  ;  and  the  second  factor  by  subtracting 
the  first  root  from  the  unknown  quantity  x;  hence, 

Every  equation  of  the  second  degree  may  be  resolved  into 
two  binomial  factors  of  the  first  degree,  the  first  terms,  in 
both  factors,  being  the  unknown  quantity,  and  the  second 
terms,  the  roots  of  the  equation,  taken  with  contrary  signs. 

162.  Into  how  many  binomial  factors  of  the  firs',  degree  may  every 
equation  of  the  second  degree  be  resolved  ?  What  a:  e  the  first  terms  cf 
these  factors  ?  What  the  second  ? 


FORMATION      OF      EQUATIONS.  229 

THIRD    PROPERTY. 

163.  It*  we  add  Equations  (4)  and  (5),  Art.  161,  we 
have, 

xf    —   —  p  +  -\/q  +  p2 
x"  =   —  p  —  -\fq  +  p2 

x'  +  x"  =   —  2p  ;  tliat  is, 

In  every  reduced  equation  of  the  second  degree,  the  sum 
of  the  two  roots  is  equal  to  the  coefficient  of  the  second  term, 
taken  with  a  contrary  sign, 

FOURTH    PROPERTY. 

164.  If  we  multiply  Equations  (4)  and  (5),  Art.  161, 
member  by  member,  we  have, 


x'  x  x      =     -  p  + 

=  p2-  (q  +p2)  =   -  q;    that  is, 
In  every  equation  of  the  second  degree,  the  product  of 
the  two  roots  is  equal  to  the  known  term  in  the  second  mem- 
ber, taken  with  a  contrary  sign. 


FORMATION    OF   EQUATIONS    OF  THE   SECOND    DEGREE. 

165.  By  taking  the  converse  of  the  second  property, 
(Art.  162),  we  can  form  equations  which  shall  have  given 
roots ;  that  is,  if  they  are  known,  we  can  find  the  corre- 
sponding equations  by  the  following 

RULE. 

I.    Subtract  each  root  from  the  unknown  quantity  : 

163.  Whai  is  the  algebraic  sum  of  the  roots  equal  to  in  every  equation 
of  the  second  degree  ? 

164.  What  is  the  product  of  the  roots  equal  to? 

165.  How  will  you  find  the  equation  when  the  roots  are  known  ? 


230  ELEMENTARY      ALGEBRA. 

II.   Multiply  the  results  together,  and  place  their  product 
equal  to  0. 

EXAMPLES. 

NOTE. — Let  the  pupil  prove,  in  every  case,  that  the  roots 
will  satisfy  the  third  and  fourth  properties. 

1.  If  the  roots  of  an  equation  are  4  and  —  5,  what  is  the 
equation  ?  Ans.  x2  +  x  =  20. 

2.  What  is  the  equation  when  the  roots  are  1  and  —  3  ? 

Ans.   x2  +  2x  =  3. 

3.  What  is  the  equation  when  the  roots  are  9  and  —  10  ? 

.  Ans.   x2  -{-  x  =  90. 

4.  What  is  the  equation  whose  roots  are  6  and  —  10? 

Ans.  x2  +  4x  =  60. 

5.  What  is  the  equation  whose  roots  are  4  and  —  3  ? 

Ans.   x2  —  x  =  12. 

6.  What  is  the  equation  whose  roots  are  10  and  —  TV  ? 

Ans.   x2  —  9TVe  =  1. 

7.  What  is  the  equation  whose  roots  are  8  and  —  2  ? 

Ans.  x2  —  Qx  =  16.  . 

8.  What  is  the  equation  whose  roots  are  16  and  —  5  ? 

Ans.  x2  —  lice  =  80. 

9.  What  is  the  equation  whose  roots  are  —  4  and  —  5  ? 

Ans.   x2  -f-  Qx  =   —  20. 

10.  What  is  the  equation  whose  roots  are  —  6  and  —  7  ? 

Ans.  x2  +  13x  =   —  42. 

1 1 .  What  is  the  equation  whose  roots  are  —  -  and  —  2  ? 

g 

Ans.  x2  -f  2$x  = • 

12.  What  is  the  equation  whose  roots  are  —  2  and  —  3  ? 

Ans.   x2  -{-  5x  =   —  6. 


NUMERICAL     VALUKS     OF     TIIK     BOOTS.       231 

13.  What  is  the  equation  whose  roots  are  4  and  3  ? 

Ans.  xz  —  7x  =   —  12. 

14.  What  is  tho  equation  whose  roots  are  12  and  2? 

Aiis.   x~  —  14a;  =   —  24, 

15.  What  is  the  equation  whose  roots  are  18  and  2? 

Ans.   x2  —  20x  =   —  36. 

16.  What  is  the  equation  whose  roots  are  14  and  3  ? 

Ans.  x2  —  l*ix  =   —  42. 

4  9 

17.  What  is  the  equation  whose  roots  are  -  and  —  -? 

Ans.  xz  -\ x  =  1, 

36 

2 

18.  What  is  the  equation  whose  roots  are  5  and ? 

3 

13  10 

Ans.   x- x  =  —  • 

3  3 

19.  What  is  the  equation  whose  roots  are  a  and  5? 

Ans.   y?  —  (a  +  b)x  =  —  ab. 

20.  What  is  the  equation  whose  roots  are  c  and  —  d? 

Ans.  x2  —  (c  —  d)  x  =  cd. 


TRINOMIAL    EQUATIONS     OF    THE    SECOND    DEGEEE. 

165.*  A  trinomial  equation  of  the  second  degree  con- 
tains three  kinds  of  terms : 

1st.  A  term  involving  the  unknown  quantity  to  the  second 
degree. 

2ct.  A  term  involving  the  unknown  quantity  to  the  first 
degree ;  and 

3d.  A  known  term.     Thus, 

x2  —  4x  —  12   =  0, 
is  a  trinomial  equation  of  the  second  degree. 


232  ELEMENTARY      ALGEBRA. 

FACrORIKG. 

165.**    What  are  the  factors  of  the  trinomial  equation, 
y?  —  4x  —  12  =  0? 

A  trinomial  equation  of  the  second  degree  may  always  be 
reduced  to  one  of  the  four  forms  (Art.  160),  by  simply  trans- 
posing the  known  term  to  the  second  member,  and  then 
solving  the  equation.  Thus,  from  the  above  equation,  we 
have, 

a3  —  4x  =   12. 

Resolving  the  equation,  we  find  the  two  roots  to  be  +6 
and  —  2 ;  therefore,  the  factors  are,  x  —  6,  and  x  4-  2 
(Art.  162). 

Since  the  sum  of  the  two  roots  is  equal  to  the  coefficient 
of  the  second  term,  taken  with  a  contrary  sign  (Art.  163) ; 
and  the  product  of  the  two  roots  is  equal  to  the  known 
term  in  the  second  member,  taken  with  a  contrary  sign,  or 
to  the  third  term  of  the  trinomial,  taken  with  the  same 
sign :  hence  it  follows,  that  any  trinomial  may  be  factored 
by  inspection,  when  two  numbers  can  be  discovered  whose 
algebraic  sum  is  equal  to  the  coefficient  of  the  second  term, 
and  whose  product  is  equal  to  the  third  term. 

EXAMPLES 

1.  What  are  the  factors  of  the  trinomial,  xz  —  9x  —  36  ? 

It  is  seen,  by  inspection,  that  —  12  and  +  3  will  fulfil  the 
conditions  of  roots.  For,  12  —  3  =  9 ;  that  is,  the  co- 
efficient of  the  second  term  with  a  contrary  sign ;  and 
12  X  —  3  =  —  36,  the  third  term  of  the  trinomial ;  hence, 
the  factors  are,  x  —  12,  and  x  4-  3. 

2.  What  are  the  factors  of  &  —  tx  -  30  =  0  ? 

Ans.   x  —  10,   and  x  +  3 


TRINOMIAL      KQDATIONS.  233 

8.  What  are  the  factors  of  x~  +  I5x  +  36   =:   0? 

Ans.   x  +  12,   and  x  +  3. 

4.  What  are  the  factors  of  x2  —  I2x  —  28  =  0  ? 

Ans.   x  —  14,   and  x  +  2. 

5.  What  are  the  factors  of  x2  —  Ix  —  8  =  0  ? 

Ans.   x  —  8,   and  x  +  1. 


TRINOMIAL   EQUATIONS   OF   THE    FORM 

x~n  -f  2pxn  —  q. 

In  the  above  equation,  the  exponent  of  &,  in  the  first  term, 
is  double  the  exponent  of  x  in  the  second  term. 

x6  —  4x3  =  32,    and    x*  +  4xz  =  117, 

are  both  equations  of  this  form,  and  may  be  solved  by  the 
rules  already  given  for  the  solution  of  equations  of  the 
second  degree. 
In  the  equation, 


we  see  that  the  first  member  will  become  a  perfect  square, 
by  adding  to  it  the  square  of  half  the  coefficient  of  xn  ;  thus, 


in  which  the  first  member  is  a  perfect  square.    Then,  ex- 
tracting the  square  root  of  both  members,  we  ha\  e, 

xn  +  p  =    ±  -\/q  +  p2 ; 
hence,  xn  —  —  p  ±  -\fq  +  p2 ; 

then,  by  taking  the  nth  root  of  both  members, 

and  x"  =    v  —  p  —  -\/—~p  +  pz> 


234  ELEMENTARY       ALGEBRA. 

EXAMPLES. 

1.  What  are  the  values  of  jc  in  the  equation, 

x6  +  Qx3  =   112? 
Completing  the  square, 

x6  +  Qx3  +  9  =   112  +  9  =  121 ; 
then,  extracting  the  square  root  of  both  members, 
x3  +  3  =   ±  -/121   =   ±  11 J  hence, 


x'  =  *f—  3  +  11,    and    x"  =  y—  3  —  11  ;  hence, 

x'  =  ^/8  =  2,  and    x"  =  */—  14  =   -- 

2.  What  are  the  values  of  a;  in  the  equation, 

x*  —  8xz  =  9  ? 
Completing  the  square,  we  have, 

x*  —  8x2  +  16  =   9  +  16   =   25. 
Extracting  the  square  root  of  both  members, 
x*—  4  =   ±  -y/25  =   ±  5  ;  hence, 


x'  =  ±  -y/4  +  5,  and  cc"  =  ±  -y/4  —  5  ;  hence, 
=  +  3  and  —  3  ;  and  a"  =  +  -/—  1  and  —  -/— 
3.  What  afe  the  values  of  x  in  the  equation, 

SB6  +  20Z3  =   69  ? 
Completing  the  square, 

a;6  +  20a:3  +  100   =  69  +  100  =  169. 
Extracting  the  square  root  of  both  members, 

a3  +  10  =    ±  yl69   =    ±  13  ;  hence, 


x'  =  \/-  10  +  13,     and    x"  =  3^/—  10  -  13. 
x'  =  3/3,     and    a"  =  ^/-  2o. 


TRINOMIAL,    EQUATIONS.  235 

4.  Wha:  are  the  values  of  x  in  the  equation, 

«*  —  2xz  =  3  ? 
Ans.  x'  =  ±  y's,  and  a"  =  ±  y'—  i. 

5.  What  are  the  values  of  x  in  the  equation, 

x*  +  8a;3  —  9  ? 

.   x'  =  1,   and  SB"  =  \~  $• 


6.  Given  x  +  -/Gee  +  4  =   12,   to  find  cc. 
Transposing  cc  to  the  second  member,  and  then  squaring, 
Qx  +  4  =  cc2  —  24a;  +  144  ; 
.-.     a;2  —  33a;  =   —140; 
and,  x'  =  28,     and    x"  =  5. 


7.   4a;  -f  4        +  2  =  7.  ^ln«.   SB'  =  4J-,   x"  — 


8.   a;  +  y5x  +10  —  8.  Ans.  x'  =  18,   ce"  =  3. 


NUMERICAL     VALUES      OF     THE     ROOTS. 

166.  We  have  seen  (Art.  160),  that  by  attributing  all 
possible  signs  to  2p  and  <?,  we  have  the  four  following 
forms : 

a3+  Zpx  =  q (1.) 

xz  —  2px  —  q (2.) 

xz  +  2px  =    —  q (3.) 

a2  —  2px  =•   —  q (4.) 

1C6.  To  how  many  forms  may  every  equation  cf  the  second  degree  be 
reduced  ?  What  are  they  ? 


236  ELEMENT  A  U^       ALGEBRA. 


First  Form. 

167.  Since    q    is  positive,   we  know,  from    Property 
Fourth,  that  the  product  of  the  roots  must  be  negative ; 
hence,  the  roots  have  contrary  signs.     Since  the  coefficient 
22)  is  positive,  we  know,  from  Property  Third,  that  the  alge- 
braic sum  of  the  roots  is  negative ;  hence,  the  negative  root 
is  numerically  the  greater. 

Second  Form. 

168.  Since  q  is  positive,  the  product  of  the  roots  must 
be  negative;  hence,  the  roots  have  contrary  signs.    Since 
2p  is  negative,  the  algebraic  sum  of  the  roots  must  be  posi- 
tive ;  hence,  the  positive  root  is  numerically  the  greater. 

Third  Form. 

169.  Since  q  is  negative,  the  product  of  the  roots  is 
positive  (Property  Fourth) ;  hence,  the  roots  have  the  same 
sign.    Since  2p  is  positive,  the  sum  of  the  roots  must  be 
negative ;  hence,  both  are  negative. 

Fourth  Form. 

170.  Since  q  is  negative,  the  product  of  the  roots  is 
positive ;  hence,  the  roots  have  the  same  sign.     Since  2p  is 
negative,  the  sum  of  the  roots  is  positive ;  hence,  the  roots 
are  both  positive. 

167.  What  sign  has  the  product  of  the  roots  in  the  first  form?     How 
are  their  signs?     Which  root  is  numerically  the  greater  ?     Why  ? 

168.  What  sign  has  the  product  of  the  roots  in  the  second  form  ?     How 
are  the  signs  of  the  roots  ?     Which  root  is  numerically  the  greater? 

169.  What  sign  has  the  product  of  the  roots  in  the  third  form  ?     How 
are  their  signs  ? 

170.  What  sign  has  the  product  of  the  roots  in  the  fourth  form  ?    How 
are  the  signs  of  the  roots  ? 


NUMERICAL     VALUE     OF     THE     ROOTS.          237 

First  and  Second  Forms. 

171.  If  we  make  q  =  0,  the  first  form  becomes, 

«2  +  2px  =  0,     or    x(x  +  2p)  =  0 ; 

which  shows  that  one  root  is  equal  to  0,  and  the  other  to  —  2jo. 

Under  the  same  supposition,  the  second  form  becomes, 

xz  —  2px  =  0,    or    x(x  —  2p)   —  0 ; 

which  shoAvs  that  one  root  is  equal  to  0,  and  the  other  to 
2p.  Both  of  these  results  are  as  they  should  be ;  since,  when 
q,  the  product  of  the  roots,  becomes  0,  one  of  the  factors 
must  be  0 ;  and  "hence,  one  root  must  be  0. 

Third  and  Fourth  Forms. 

172.  If,  in  the  Third  and  Fourth  Forms,  q>pz,  the 
quantity  under  the  radical  sign  will  become  negative  ;  hence, 
its  square  root  cannot  be  extracted  (Art.  137).    Under  this 
supposition,  the  values  of  x  are  imaginary.    How  are  these 
results  to  be  interpreted  ? 

If  a  given  number  be  divided  into  two  parts,  their  pro- 
duct  will  be  the  greatest  possible,  when  the  parts  are  equal. 

Denote  the  number  by  2p,  and  the  difference  of  the  parts 
by  d',  then, 

p  +  -  ==    the  greater  part,     (Page  120.) 
2 


and, 
and, 

p- 

P2- 

2 
dz 

4 

the  less  part, 
P,  their  product. 

171.  If  we  make  q  =  0,  to  what  does  the  first  form  reduce?  What, 
then,  are  its  roots  ?  Under  the  same  supposition,  to  what  does  the  second 
form  reduce  ?  What  are,  then,  its  roots? 

1*72.  If  q  >  pa,  in  the  third  and  fourth  forms,  what  takes  place? 

If  a  number  be  divided  into  two  parts,  when  will  the  product  be  the 
greatest  possible  ? 


238  ELEMENTARY      ALGEBRA 

It  is  plain,  that  the  product  P  will  increase,  as  d  dimin- 
ishes, and  that  it  Avill  be  the  greatest  possible  when  d  •=  0  ; 
for  then  there  will  be  no  negative  quantity  to  be  subtracted 
from  p2,  in  the  first  member  of  the  equation.  But  when 
d  =  0,  the  parts  are  equal ;  hence,  the  product  of  the  two 
parts  is  the  greatest  when  they  are  equal. 

In  the  equations, 

x2  +  2px  =  —  <7,       x2  —  2px  =   —  q, 

2p  is  the  sum  of  the  roots,  and  —  q  their  product;  and 
hence,  by  the  principle  just  established,  the  product  q, 
can  never  be  greater  than  p7.  This  condition  fixes  a  limit 
to  the  value  of  q.  If,  then,  we  make  q  >  ^>2,  we  pass  thie 
limit,  and  express,  by  the  equation,  a  condition  which  cannot 
be  fulfilled ;  and  this  incompatibility  of  the  conditions  is 
made  apparent  by  the  values  of  x  becoming  imaginary. 
Hence,  we  conclude  that, 

When  the  values  of  the  unknown  quantity  are  imaginary, 
the  conditions  of  the  proposition  are  incompatible  with 
each  other. 

EXAMPLES. 

1.  Find  two  numbers,  whose  sum  shall  be  12  and  pro 
duct  46. 

Let  x  and  y  be  the  numbers.  . 

By  the  1st  condition,  x  +  y  =  12  ; 
and  by  the  2d,  xy  =  46. 

The  first  eqiiation  gives, 

x  =  12  —  y. 
Substituting  this  value  for  x  in  the  second,  we  have, 

12y  -  y2  =  46  ; 

and  changing  the  signs  of  the  terms,  we  have, 
»  _  12     =  -  46 


NUMERICAL     VALUE     OF     THE     KOOTS.         239 

Then,  by  completing  the  square, 

2/3  —  12y  +  36  =   —46  +  36  =   —  10; 
which  gives,  y'    =  6  +  ^/—  10, 

and,  y"  =  6  —  •/—  10 ; 

both  of  which  values  are  imaginary,  as  indeed  they  should 
be,  since  the  conditions  are  incompatible. 

2.  The  sum  of  two  numbers  is  8,  and  their  product  20: 
what  are  the  numbers  ? 

Denote  the  numbers  by  x  and  y. 
By  the  first  condition, 

x  +  y  =  8; 

and  by  the  second,  xy  =  20. 

The  first  equation  gives, 

x  =  8  —  y. 
Substituting  this  value  of  x  in  the  second,  we  have, 

8y  -  y2  =  20  ; 
changing  the  signs,  and  completing  the  square,  we  have, 

2/2  -  8y  +  16  =   -  4 ; 
and  by  extracting  the  root, 

y'  =  4  +  •>/—  4,    and    y"  =  4  —  y' —  4. 
These  values  of  y  may  be  put  under  the  forms  (Art.  142), 
y  =  4  -f-  2-v/— ^j    and    y  =  4  —  2^—  1. 

3    What  are  the  values  of  x  in  the  equation, 
x*       2x  =   —  10? 


(  85'     rr     —1+3   <J  —   1. 

yl».<?.  -j  '. 

(  ic"    =r:     —    1    -    3V—   1. 


240  ELEMENTARY      ALGEBKA. 


PEOBLEMS. 

1.  Find  a  number  such,  that  twice  its  square,  added  to 
three  times  the  number,  shall  give  65. 

Let  x  denote  the  unknown  number.    Then,  the  equation 
of  the  problem  is, 

2«2  +  3x  =  65  ; 
whence, 


3 

/65         9 

3        23 

X 

=    ~4±\ 

'T  +  fi  = 

-[-  

4  3-    4 

Therefore, 

,              3 
X    —    —  - 

,    23   _ 

and    «"  = 

3        23 

13 

4 

~  4  ~  T  : 

"  ~2~ 

Both  these  values  satisfy  the  equation  of  the  problem. 
For, 

2  X  (5)2  +3x5   =   2x25  +  15   =  65; 

13\2  ,                   13         169        39         130 
-  ^r)  +  3  X  -  —   =— —  =  — -   =65. 


NOTES. — 1.  If  we  restrict  the  enunciation  of  the  problem 
to -its  arithmetical  sense,  in  which  "added"  means  numer- 
ical increase,  the  first  value  of  x  only  will  satisfy  the  con- 
ditions of  the  problem. 

2.  If  we  give  to  "  added,"  its  algebraical  signification 
(when  it  may  mean  subtraction  as  well  as  addition),  the 
problem  may  be  thus  stated : 

To  find  a  number  such,  that  twice  its  square  diminished 
by  three  times  the  number,  shall  give  65. 

The  second  value  of  x  will  satisfy  this  enunciation ;  for, 


/13V  13 

HT)  - 3  x  T  = 


169       39 

-r  ~  T  =  66- 


P  K  O  B  L  E  M  8 . 


241 


3.  The  root  which  results  from  giving  the  plus  sign  to  the 
radical,  is,  generally,  an  answer  to  the  question  in  its  arith- 
metical sense.  The  second  root  generally  satisfies  the  pro- 
blem under  a  modified  statement. 

Thus,  in  the  example,  it  was  required  to  find  a  number, 
of  which  twice  the  square,  added  to  three  times  the  num- 
ber, shall  give  65.  Now,  in  the  arithmetical  sense,  added 
means  increased ;  but  in  the  algebraic  sense,  it  implies  dimi- 
nution when  the  quantity  added  is  negative.  In  this  sense, 
the  second  root  satisfies  the  enunciation. 

2.  A  certain  person  purchased  a  number  of  yards  of  cloth 
for  240  cents.  If  he  had  purchased  3  yards  less  of  the  same 
cloth  for  the  same  sum,  it  would  have  cost  him  4  cents  more 
per  yard  :  how  many  yards  did  he  buy  ? 

Let  x    denote  the  number  of  yards  purchased. 

240 
Then,   -    -   will  denote  the  price  per  yard. 

•B 

If,  for  240  cents,  he  had  purchased  three  yards  less,  that 
•s,  x  —  3  yards,  the  price  per  yard,  under  this  hypothesis, 

240 

would  have  been  denoted  by  -  — -  •  But,  by  the  condi- 
tions, this  last  cost  must  exceed  the  first  by  4  cents.  There- 
fore, we  have  the  equation, 

240          240 
x  —  3         x  ' 


whence,  b 
and, 
therefore, 

y  reducing 

r.i       — 

y~ 

itij 

3x  = 

180, 
3  -t  21 

N/!  +  > 

15,    and 

RCl 

~   2  ± 
x'  = 

oU    — 

x"  = 

2 
-  12. 

NOTES. — 1.  The  value,  x'  =  15,  satisfies  the  c  nunciation 
in  its  arithmetical  sense.     For,  if  15  yards  cost  240  cents, 
11 


242  ELEMENTARY     ALGEBKA. 

240  -4-15  =  16  cents,  the  price  of  1  yard  ;  and  240  -f-  12  =  20 
cents,  the  price  of  1  yard  under  the  second  supposition. 

2.  The  second  value  of  x  is  an  answer  to  the  following 
Problem : 

A  certain  person  purchased  a  number  of  yards  of  cloth 
for  240  cents.  If  he  had  paid  the  same  for  three  yards  more, 
it  would  have  cost  him  4  cents  less  per  yard :  how  many 
yards  did  he  buy  ? 

This  would  give  the  equation  of  condition, 
240          240 

-i^'iTi  =  4;  or' 

xz  —  3x  =  180; 
the  same  equation  as  found  before ;  hence, 

A  single  equation  will  often  state  two  or  more  arith- 
metical problems. 

This  arises  from  the  fact  that  the  language  of  Algebra  is 
more  comprehensive  than  that  of  Arithmetic. 

3.  A  man  having  bought  a  horse,  sold  it  for  $24.    At  the 
sale  he  lost  as  much  per  cent,  on  the  price  of  the  horse,  as 
the  horse  cost  him  dollars :  what  did  he  pay  for  the  horse  ? 

Let  x  denote  the  number  of  dollars  that  he  paid  for  the 
horse.  Then,  x  —  24  will  denote  the  loss  he  sustained.  But 

£C 

as  he  lost  x  per  cent,  by  the  sale,  he  must  have  lost  — — 

upon  each  dollar,  and  upon  x  dollars  he  lost  a  sum  denoted 

xz 
b\     :   we  have,  then,  the  equation, 

*      100 

— -  =  x  —  24;    whence,    a3  —  lOOa  z=   —  2400, 
100 


PROBLEMS.  il4:i) 


and,  x  =   50  ±  -v/2500  —  2400   =   50  ±  10. 

Therefore,  x'  =  60,     and    «"  =  40. 

Both  of  these  roots  will  satisfy  the  problem. 

For,  if  the  man  gave  $60  for  the  horse,  and  sold  him  for 
$24,  he  lost  $36.  From  the  enunciation,  he  should  have  lost, 
60  per  cent,  of  $60  ;  that  is, 

60  60  X  60 

100  °f6°          -loo"  >6' 

therefore,  $60  satisfies  the  enunciation. 

Had  he  paid  $40  for  the  horse,  he  would  have  lost  by  the 
sale,  $16.  From  the  enunciation,  he  should  have  lost  40  per 
cent,  of  $40 ;  that  is, 

40  40  X  40 

of  40  = =  16  : 

100  100 

therefore,  $40  satisfies  the  enunciation. 

4.  The  sum  of  two  numbers  is  11,  and  the  sum  of  their 
squares  is  61 :  what  are  the  numbers?  Ans,   5  and  6. 

5.  The  difference  of  two  numbers  is  3,  and  the  sum  of  their 
squares  is  89  :  what  are  the  numbers  ?  Ans.    5  and  8. 

6.  A  grazier  bought  as  many  sheep  as  cost  him  £60,  and 
after  reserving  fifteen  out  of  the  number,  he  sold  the  re- 
mainder for  £54,  and  gained  25.  a  head  on  those  he  sold : 
how  many  did  he  buy  ?  Ans.    75. 

7.  A  merchant  bought  cloth,  for  which  he  paid  £33  15s., 
which  he  sold  again  at  £2  8s.  per  piece,  and  gamed  by  the 
bargain  as  much  as  one  piece  cost  him :  how  many  pieces 
did  he  buy?  Ans.    15. 

8.  The  difference  of  two  numbers  is  9,  and  their  sum, 
multiplied  by  the  greater,  is  equal  to  266 :  what  are  the 
numbers?  Ans.    14  and  5. 


24-i  ELEMENTARY       A  L  G  E  B  K  A  . 

9.  To  find  a  number,  such  that  if  you  subtract  it  from  10, 
and  multiply  the  remainder  by  the  number  itself,  the  pro- 
duct will  be  21.  Ans.    7  or  3. 

10.  A  person  traveled  105  miles.     If  he  had  traveled  2 
miles  an  hour  slower,  he  would  have  been  6  hours  longer  in 
completing  the  same  distance :  how  many  miles  did  he  travel 
per  hour  ?  Ans.    7  miles. 

11.  A  person  purchased  a  number  of  sheep,  for  which  he 
paid  $224.    Had  he  paid  for  each  twice  as  much,  plus  2  dol- 
lars, the  number  bought  would  have  been  denoted  by  twice 
what  was  paid  for  each  :  how  many  sheep  were  purchased  ? 

Ans.    32. 

12.  The  difference  of  two  numbers  is  7,  and  their   sum 
multiplied  by  the  greater,  is  equal  to  130:  what  are  the 
numbers?  Ans.    10  and  3. 

13.  Divide  100  into  two  such  parts,  that  the  sum  of  their 
squares  shall  be  5392.  Ans.   64  and  38. 

14.  Two  square  courts  are  paved  with  stones  a  foot  square ; 
the  larger  court  is  12  feet  larger  than  the  smaller  one,  and 
the  number  of  stones  in  both  pavements  is  2120  :  how  long 
is  the  smaller  pavement?  Ans.    26  feet. 

15.  Two  hundred  and  forty  dollars  are  equally  distributed 
among  a  certain  number  of  persons.     The  same  sum  is  again 
distributed  amongst  a  number  greater  by  4.     In  the  latter 
case  each  receives  10  dollars  less  than  in  the  former:  how 
many  persons  were  there  in  each  case.  Ans.   8  and  12. 

16.  Two  partners,  A  and  _B,  gained  360  dollars.     A1  a 
money  was  in  trade  12  months,  and  he  received,  for  prin- 
cipal and  profit,  520  dollars.     JPs  money  was  600  dollars, 
nnd  was  in  trade  16  months:  how  much  capital  had  A  ? 

Ans.   400  dollars. 


MOKE  THAN  UNK  UNKNOWN  QUANTITY.   2-i;r> 


EQUATIONS    INVOLVING   ilOKE  THAN  ONE    UNKNOWN  QUANTITY. 

173.  Two  simultaneous  equations,  each  of  the  second 
degree,  and  containing  two  unknown  quantities,  will,  when 
combined,  generally  give  rise  to  an  equation  of  the  fourth 
degree.  Hence,  only  particular  cases  of  such  equations  can 
be  solved  by  the  methods  already  given. 

FIRST. 

Two  simultaneous  equations,  involving  two  unknown 
quantities,  can  always  be  solved  when  one  is  of  the  first 
and  the  other  of  the  second  degree. 

EXAMPLES. 

1.  Given      ,        ,      ,  ^  r     to  find  x  and  y. 

(  xz  +  y  =  100  ) 

By  transposing  y  in  the  first  equation,  we  have, 

x  =   14  -  y; 
and  by  squaring  both  members, 

x2  =   190  —  28y  +  y2. 

Substituting  this  value  for  x2  in  the  second  equation,  we 
have, 

196  —  28y  +  y2  +  y2  =  100  ; 

from  Avhich  we  have, 


By  completing  the  square, 

y2  -  14y  +  49  =  1  ; 

173.  When  may  two  simultaneous  equations  of  the  second  degree  be 
solved  ? 


246  ELEMENTARY       ALGEBRA. 

and  by  extracting  the  square  root, 

y  —  7  —   ±  -v/1   =   +  1>    and     —  1  ; 
hence,  y'  —  7  +  1   =  8,     and    y"  —  1  — .  1   =  6. 

If  we  -take  the  greater  value,  we  find  x  =  6  ;  and  if  we 
fake  the  lesser,  we  find  x  =  8. 

(  v'  —  s     »•"  —   fi 

j  1     »'.         —       U»        *o  — •       U* 

AnS'\y>=   6,    y"^   8 

VERIFICATION. 

For  the  greater  value,  y  =  8,  the  equation, 

a;   +  y    =     14,     gives      6  +    8  —  14 ; 

and,       a2  +  y2  =  100,     gives     36  +  64  =  100. 
For  the  value  y  —  6,  the  equation, 

x   +  y   =     14,     gives       8  +    6  =  14 ; 

and,       a8  +  y2  =  100,     gives    64  +  36   =  100. 

Hence,  both  sets  of  values  satisfy  the  given  equation. 

2.  Given   j  [  to  find  x  and  y. 

(  xz  —  y2  =  45  ) 

Transposing  y  in  the  first  equation,  we  have, 

x  =  3  +  y; 
then,  squaring  both  members, 

a;2  =  9  +  6y  +  y2. 

Substituting  this  value  for  a;2,  in  the  second  equation,  we 

have, 

9  +  6y  +  y2  —  y2  =  45 ; 

whence,  we  have, 

6y  =  36,    and    y    =•  6. 


SIMULTANEOUS      EQUATIONS.  247 

Substituting  this  value  of  y,  in  the  first  equation,  we  have, 

x  —  6  =  3, 
and,  consequently,     x'  =  3  +  6  =  9. 

VERIFICATION. 

x  —  y    =     3,    gives      9  —    6  =     8 ; 
and,  x2—  y2  =  45,    gives    81  —  36  =  45. 

Solve  the  following  simultaneous  equations : 

+ y    =  12  r  «''=  7. 


g 

'  =     5. 

ix  —  y    =       3)  .         (  a;'  =  9,    x"=   —  6. 

4.  J       2,0  ,  ,  H    J-  -A7W.     -I        ,  „ 

(  a;2+  y2  =z  117  f  (  y'  =  6,    y"=   —  9. 

r  />•  4.  «/   —   Q  \  /•/>•'  *j        " «; 

_4^-ry   —   y  •  A          i  x    —   o,          —  o. 

5.  4     ,  ,         ,  L         -4^5.   4     .  .. 

(  a;- —  2xy  -\-  y2  =   1  $  )  y    =  4,    y   ==  4. 

raj-y  =  5  i 

(  a$2+  2a?y  +  y2  =  225  [ 

yf  =  10,    x"=   -    5. 

y'  =     5,    y"=   -  10. 


SECOND. 


174.  Two  simultaneous  equations  of  the  second  degree, 
which  are  homogeneous  with  respect  to  the  unknown  quan- 
tityr,  can  always  be  solved. 


EXAMPLES. 


r.          j  a?  +  Sxy  =  22  (1.) 

'    (a?+  3a:y  +  2y2  =  40 (2.) 

to  find  x  and  y. 

174.  When  may  two  simultaneous  equations  of  the  second  degree  be 
solved  ? 


2i8  ELEMENTARY      ALGEBRA. 

Assume  x  =  ty,  t  being  any  auxiliary  unknown  quantity. 
Substituting  this  value  of  x  in  Equations  (  1  )  and  (  2  ), 
we  have, 

<V  +  atf  =  22,        .-.    3,2  =  fTqrisJ  (3-> 

40 

«y+  3ty*+w  =  40,      .-.   7/2  =  2  ;    (*0 


22  40 

hence, 


t2  +  3t        t*  +  3t  +  2  ' 
hence,  22£2  +  66J  +  44  =  40f2  +  120«; 

22 

reducing,  <2  _j_  3^  _  —  . 

y 

2                               11 
whence,  «'  =  -,   and   t"= 

Substituting  either  of  these  values  in  Equations  ( 3 )  or 
( 4  ),  we  find, 

y'  —   +3,    and    y"  —   —  3 

Substituting  the  plus  value  of  y,  in  Equation  ( 1 ),  we 

have, 

y?  +  9x  =  22  ; 

from  which  we  find, 

x'  =   +  2,    and    x"  =   —  11. 

If  we  take  the  negative  value,    y"  =    —  3,   we  have, 
from  Equation  ( 1 ), 

xz  —  9«  =  22  ; 
from  which  we  find, 

x'   =   +  11,    and    x"   =   —  2. 

VERIFICATION. 

For  the  values    y'  =    +3,   and  x'  =   +2,   the  giveii 

equation, 

*2  4-  3ay  =  22, 


SIMULTANEOUS      EQUATIvNS.  249 

gives,  22 +3X2X3   =   4+18   =   22; 

and  for  the  second  value,   x"  •—   —  11,   the  same  equation, 

x2  +  Zxy  =  22, 

» 
gives,     (—  11)2+  3  X  -  11  X  3   =   121  -  99  =  22. 

If,  now,  we  take  the  second  value  of  y,  that  is,  y"  =  —  3, 
and  the  corresponding  values  of  ic,  viz.,  x'  =  +11,  and 
x"  =  —  2;  for  xf  =  +11,  the  given  equation, 

x2  +  Sxy  =  22, 

gives,       II2  +  3  x  11  x  —  3   =   121  —  99  =  22; 
and  for    x"  =   —  2,    the  same  equation, 

xz  +  3xy  =  22, 
gives,     (—  2)2  +  3  X  -  2  X  —  3   ^4  +  18   =   22. 

The  verifications  could  be  made  in  the  same  way  by  em- 
ploying Equation  (  2  ). 

NOTE. — In  equations  similar  to  the  above,  we  generally 
find-but  a  single  pair  of  values,  corresponding  to  the  values 
hi  this  equation,  of  y'  =  +3,  and  x'  —  +2. 

The  complete  solution  would  give  four  pairs  of  values. ' 


X2  -       y2  =    _  9  j  j_.  -|»  »  *• 

y  =  5. 

a;  =  6. 

-<9L7^.    -{ 

(  y  =  7. 
=    470 


I    .o      —        «       —  —    a    I                                                                          (23     —     4» 

2.  .J    „  _  t                           4n».  J 

/*  -  xy    =  5  f                                          (  y  =  5. 

3.  V?  7  ^o  .  i                              -4»w. 

rz:  HO 


4.    ^  T         "•7    ' 


5   —    xy 

(  5xy  —  3y2          —  32  . 

1    i  -L.     2       o  ,  5-  -4*1'*' 

11* 


250                      ELEMENTARY      ALGEBRA. 
THIRD. PARTICULAR   CASES. 

175.  Many  other  equations  of  the  second  degree  may  be 
so  transformed,  as  to  be  brought  under  the  rules  of  solution 
already  given.  The  seven  following  formulas  will  aid  in 
such  transformation. 

(1.) 
When  the  sum  and  difference  are  known: 

x  +  y  =  s 
x  —  y  —  d. 

Then,  page  132,  Example  3, 

s  +  d        1,1,  s  —  d        1        1  , 

x  =  _-  =  -s  +  -d,     and     y  =  -^-  =  -s  - -d. 

(2.) 
When  the  sum  and  product  are  known: 

x  -{-    y  =     s (l.) 

xy  =    p (2.) 

»2  +  2#y  +  y2  =     s2,    by  squaring  ( 1 ) ; 
4xy  =  4p,  by  mult.  ( 2 )  by  4. 

K2  —  2xy  +  yz  =  s2  —  4/>,    by  subtraction. 

x  —  y  =   ±  yV  —  4p,    by  ext.  root. 
But,  x  +  y  —  s\ 

hence 


and, 


I7i5.  What  is  the  first  formula  of  this  article?    What  the  second? 
Third?    Fourth?    Fifth?    Sixth?    Seventh? 


SIMULTANEOUS      EQUATIONS. 


251 


(3.) 
When  the  difference  and  product  are  known: 

«  -    y  =     d  .......  .     (  1.) 

xy  =  p  .    .    %  f    .    .    .     (2.) 
x2  —  2xy  +  y2  =     of2,    by  squaring  (  1  )  • 

4xy  =  4p,     mult.  (  2  )  by  4. 

«2  -f  2;cy  +  y7  =     <?2  +  4p,    by  adding. 

a;  +       =  ± 


4.) 

When  the  sum  of  the  squares  and  product  are  known  . 
a2  +  2,2=  «..(!.)     ay=i>..(2.)    .-.  2ay  =  2^>  .  .  (3.) 

Adding  (  1  )  and  (  3  ),  x2  +  2«y  +  y2  =  s  +  2p  ; 
hence,  SB  +  y=   ±  yT+~2^     (4.) 

Subtracthig  (3)  from  (  1  ),    xz  —  2xy  +  y2  =  5  —  2p  ; 
hence,  x  —  y  =    ±  •/«  —  2p     (5.) 

Combining  (4)  and  (5),  x  =  i-y/s  +  2p  +  %-Js  —  2p, 


(5.) 
When  the  sum  and  sum  of  the  squares  are  known  : 

x   +  y    =  s        .....     (  1.) 
cc2  +  y2  =  s'       .....     (2.) 

x2  +  2xy  +  y2  =  s2    by  squaring  (  I  ) 


2cey  =  sz  —  s' 

s2-  s' 

*y  =  —7T- 


=  p. 


(3-) 


ELEMENTABY      ALGEBKA. 

By  putting  xy  =  p,  and  combining  Equations  (  1  )  and 
(3  ),  by  Formula  (2),  we  find  the  values  of  x  and  y. 

(6.) 
"When  the  sum  and  sum  of  the  cubes  are  known  : 

x  +  y   =  8        ....     (1.) 
3.3  _|_  y3  —  i52     .     .     .     .     (2.) 

JB3  +  3x2y  +  3xyz  +  y3  =  512     by  cubing  (  1  ). 


3a:2y  +  3xy2  =360  by  subtraction. 

3xy(x  +  y)   =   360  by  factoring. 

3ay(8)   =  360  from  Equa.  (  1  ). 

24xy  =  360 

hence,  xy  —  15  ....     (8.) 

Combining  (  1  )  and  (  3  ),  we  find  a;  =  5  and  y  =  3 

|»0 

When  we  have  an  equation  of  the  form, 

(»  +  yY  +  (x  +  y)   =  q. 
Let  us  assume  x  4-  y  =  2. 
Then  the  given  equation  becomes, 


+  2  =  ?;    and    s  =   -  -  ±  \    q  +  - 


EXAMPLES. 


2  t  ~i  \  \ 

ir  I 

1.  Given   •{  cc  +  y  +  z   =    7  (  2  )>  to  find  a,  y,  and  as, 

'  +   2,2+22  =   21  (3)) 


SIMULTANEOUS      EQUATIONS.  253 

Transposing  y  in  Equation  (  2 ),  we  have, 

x  +  z  =  7-y;         ...    (4.) 
then,  squaring  the  members,  we  have, 

K2  -f  2cez  +  z2  =  49  —  14y  +  y2. 

If  now  we  substitute  for  2a:z,  its  value  taken  from  Equa- 
tion ( 1 ),  we  have, 

a;2  +  2y2  +  z2  =  49  —  14y  -f  y2 ; 
and  cancelling  y2,  in  each  member,  there  results, 
«2  -f  y2  +  z2  =  49  —  14  y. 

But,  from  Equation  ( 3 ),  we  see  that  each  member  of  the 
last  equation  is  equal  to  21 ;  hence, 

49  —  14y  =  21, 
and,  14y  =  49  —  21   —  28 , 

28 
hence,  y  =  —  =  2. 

Substituting  this  value  of  y  in  Equation  (1 ),  gives, 

xz  =  4 ; 
and  substituting  it  in  Equation  (4  ),•  gives, 

x  +  z  =  5,    or    a;  =  5  —  z. 

Substituting  this  value  of  «,  in  the  previous  equation,  we 

obtain 

5z  -  zz  =4,     or    zz  —  fiz  =   —  4 ; 

and  by  completing  the  square,  we  have, 

22  -  5z  +  6.25   =  2.5. 

and,       2  —  2.5   =    ±  \/2^  =   +  1.5,     or     —  1.5; 
hence,  z  =  2.5  -f  1.5  =  4,   and   2  =   -f  2.5  —  1.6   =  1 


254  ELEMENTARi'      ALGEBRA. 

_  \ 

2.  Given    x   +  yxy  +  y    =     19  )•         ~    ,  , 

,      ,     •  ,  „  f   to  find  x  and  y. 

and    x2  +      ay  +  y2  =  133  ) 

Dividing  the  second  equation  by  the  first,  AVC  have, 

x  —  i/xy  +    y  =     7 
but,  a;  +  Va-y  +    y  =  19 


hence,  by  addition,  2«  +  2y  =  26 

or,  a;  +    y  =  13 

and  substituting  in  1st  Equa.,  -v/ay  +  13  =  19 
or,  by  transposing,  V^y  —     ^ 

and  by  squaring,  ay  =  36. 

Equation  2d,  is  a2  +  xy  +  y2  =  133 

and  from  the  last,  we  have,  3a;y  =108 

Subtracting,  x2  —  2xy  +  y2  =     25 

hence,  x  —  y  =  db  5 

but,  x  +  y  =     13 

hence,      x  =  9   and   4  ;     and    y  =  4   and   9. 


PROBLEMS. 

1.  Find  two  numbers,  such  that  their  sum  shall  be  15  and 
the  sum  of  their  squares  113. 

Let  x  and  y  denote  the  numbers ;  then, 
x  4-  y  =  15,     (1.)        and    a2  +  y2  =  113.    (2.) 
From  Equation  ( 1 ),  \ve  have, 

a;2  =  225  —  30y  -f  y2 
Substituting  this  value  in  Equation  (  2  ), 

225  -80y  +  y2  +  y2  =   113; 


PROBLEMS.  256 

hence,  2y2  —  30y  =   —  112; 

2/2  -  I5y  =   -    56,  . 
hence,  '  y'  =.  8,     and    y"  —  7. 

The  first  value  of  y  being  substituted  in  Equation  ( 1 ), 
gives  x'  =  7  ;  and  the  second,  x"  =  8.  Hence,  the  num- 
bers are  7  and  8. 

2.  To  find  two  numbers,  such  that  their  product  added  to 
their  sum  shall  be  1 7,  and  their  sum  taken  from  the  sum  of 
their  squares  shall  leave  22. 

Let  x  and  y  denote  the  numbers;  then,  from  the  con- 
ditions, 

(x  +  y)  +  xy  =  17.        ...     (1.) 

x*  +  y>  -  (x  +  y)   =  22.        .     .     .     (2.) 
Multiplying  Equation  ( 1 )  by  2,  we  have, 

<2xy  +  2(x  +  y)   =  34.        ...     (3.) 
Adding  ( 2 )  and  ( 3 ),  we  have, 

a2  +  Ixy  +  y2  +  (a?  +  y)  =  56 ; 
hence,  (x  +  y}z  +  (x  +  y)  -  66.      .     .    (4.) 

Regarding  (x  -\-  y)  as  a  single  unknown  quantity  (page 
248), 


Substituting  this  value  in  Equation  ( 1 ),  we  have, 

7  +  xy  =  17,      and     y  =  5. 
Hence,  the  numbers  are  2  and  5. 

3.  What  two  numbers  are  those  whose  sum  is  8,  and  sum 
of  their  squares  34  ?  Ans.  5  and  3. 


256  ELEMENTARY      ALGEBIIA. 

4.  It  is  required  to  find  two  such  numbers,  that  the  first 
shall  be  to  the  second  as  the  second  is  to  16,  and  the  sum  oi 
whose  squares  shall  be  225  ?  Ans.    9  and  12. 

5.  "What  two  numbers  are  those  which  are  to  each  othei 
as  3  to  5,  and  whose  squares  added  together  make  1666  ? 

Ans.    21  and  35. 

6.  There  are  two  numbers  whose  difference  is  7,  and  half 
their  product  plus  30  is  equal  to  the  square  of  the  less 
number:  what  are  the  numbers?  Ans.    12  and  19. 

7.  What  two  numbers  are  those  whose  sum  is  5,  and  the 

sum  of  their  cubes  35  ?  Ans.    2  and  3. 

* 

8.  What  two  numbers  are  those  whose  sum  is  to  the 
greater  as  11  to  7,  and  the  difference  of  whose  squares  is 
132?  Ans.    14  and  8. 

9.  Divide  the  number  100  into  two  such  parts,  that  the 
product  may  be  to  the  sum  of  their  squares  as  6  to  13. 

Ans.    40  and  60. 

10.  Two  persons,  A  and  _B,  departed  from  different  places 
at  the  same  time,  and  traveled  towards  each  other.     On 
meeting,  it  appeared  that  A  had  traveled  18  miles  more 
than  £ ;  and  that  A  could  have  gone  I$'s  journey  in  15£ 
days,  but  JB  would  have  been  28  days  in  performing  A's 

journey :  how  far  did  each  travel  ?  $  A,  72  miles. 

Ans.  j     '          . 

(  J3,  54  miles. 

11.  There  are  two  numbers  whose  difference  is  15,  and 
half  their  product  is  equal  to  the  cube  of  the  lesser  number : 
what  are  those  numbers  ?  Ans.   3  and  18. 

12.  What  two  numbers  are  those  whose  sum,  multiplied 
by  the  greater,  is  equal  to  77  ;  and  whose  difference,  multi- 
plied by  the  less,  is  equal  to  12  ? 

Am.   4  and  7,  or  f  \/2  and  y  V^- 


PEOBLKMS.  257 

13.  Divide  100  into  two  such  parts,  that  the  sum  of  their 
square  roots  may  be  14.  Ans.   64  and  36. 

14.  It  is  required  to  divide  the  number  24  into  two  such 
parts,  that  their  product  may  be  equal  to  35  times  their  dif- 
ference. Ans.    10  and  14. 

15.  The  sum  of  two  numbers  is  8,  and  the  sum  of  their 
cubes  is  152  :  what  are  the  numbers  ?  Ans.   3  and  5. 

16.  Two  merchants  each  sold  the  same  kind  of  stuff;  the 
second  sold  3  yards  more  of  it  than  the  first,  and  together 
they  receive  35  dollars.     The  first  said  to  the  second,  "  I 
would  have  received  24  dollars  for  your  stuff;"  the  other 
replied,  "And  I  should  have  received  12^  dollars  for  yours  :" 
how  many  yards  did  each  of  them  sell  ? 

1st  merchant  x'  =  15,  x"  =  5. 


17.  A  widow  possessed  13,000  dollars,  which  she  divided 
into  two  parts,  and  placed  them  at  interest  in  such  a  manner 
that  the  incomes  from  them  were  equal.    If  she  had  put  out 
the  first  portion  at  the  same  rate  as  the  second,  she  would 
have  drawn  for  this  part  360  dollars  interest  ;  and  if  she 
had  placed  the  second  outsat  the  same  rate  as  the  first,  she 
would  have  drawn  for  it  490  dollars  interest  :  what  Avere 
the  two  rates  of  interest  ?  Ans.   7  and  6  per  cent. 

18.  Find  three  numbers,  such  that  the  difference  between 
the  third  and  second  shall  exceed  the  difference  between  the 

•  second  and  first  by  6  ;  that  the  sum  of  the  numbers  shall  be 
83,  and  the  sum  of  their  squares  467. 

Ans.   5,  9,  and  19. 

19.  "What  number  is  that  which,  being  divided  by  the 
product  of  its  two  digits,  the  quotient  will  be  3  ;  and  if  1  8 
be  added  to  i$  the  resulting  number  will  be  expressed  by 
the  digits  inverted?  Ans.   24. 


258 


ELEMENTARY        A  L  G  E  Ji  K  A . 


20.  What  two  numbers  are  those  which  are  to  each  other 
as  m  to  n,  and  the  sum  of  whose  squares  is  b  ? 

myb  n-\fb 


Ans. 


21.  What  t\\ro  numbers  are  those  which  are  to  each  other 
as  m  to  ft,  and  the  difference  of  whose  squares  is  b  ? 

Ans.   — —  — . - , 


22.  Required  to  find  three  numbers,  such  that  the  product 
of  the  first  and  second  shall  be  equal  to  2  ;  the  product  of 
the  first  and  third  equal  to  4,  and  the  sum  of  the  squares 
of  the  second  and  third  equal  to  20.  Ans.    1,  2,  and  4. 

23.  It  is  required  to  find  three  numbers,  whose  sum  shall 
be  38,  the  sum  of  their  squares  634,  and  the  difference 
between  the  sQcond  and  first  greater  by  7  than  the  difference 
between  the  third  and  second.  Ans.    3,  15,  and  20. 

24.  Required  to  find  three  numbers,  such  that  the  product 
of  the  first  and  second  shall  be  equal  to  a ;  the  product  of 
the  first  and  third  equal  to  b ;  and  the  sum  of  the  squares 
of  the  second  and  third  equal  to  c. 


Ans. 


x  =. 


y  = 


25.  What  two  numbers  are  those,  whose  sum,  multiplied 
by  the  greater,  gives  144;  and  whose  difference,  multiplied 


by  the  less,  gives  14  ? 


Ans.    9  and  7. 


PROPORTIONS      AND      PROGRESSIONS.      259 


CHAPTER  IX. 

OF  PROPORTIONS  AND  PROGRESSIONS. 

176.  Two  quantities  of  the  same  kind  may  be  compared, 
the  one  with  the  other,  in  two  ways : 

1st.  By  considering  how  much  one  is  greater  or  less  than 
the  other,  which  is  shown  by  their  difference ;  and, 

2d.  By  considering  hoic  many  times  one  is  greater  or  less 
than  the  other,  which  is  shown  by  their  quotient. 

Thus,  in  comparing  the  numbers  3  and  12  together,  with 
respect  to  their  difference,  we  find  that  12  exceeds  3,  by  9; 
and  in  comparing  them  together  with  respect  to  their  quo- 
tient, we  find  that  12  contains  3,  four  times,  or  that  12  is  4 
tunes  as  great  as  3. 

The  first  of  these  methods  of  comparison  is  called  Arith- 
metical Proportion,  and  the  second,  Geometrical  Propor- 
tion. 

Hence,  Arithmetical  Proportion  considers  the  relation  of 
quantities  with  respect  to  their  difference,  and  Geometrical 
Proportion  the  relation  of  quantities  with  respect  to  their 
quotient. 

176.  In  how  many  ways  may  two  quantities  be  compared  the  one  with 
the  other?  What  does  the  first  method  consider?  What  the  second? 
What  is  the  first  of  these  methods  called?  What  is  the  second  called? 
How  then  do  you  define  the  two  proportions? 


260  ELEMENTARY      ALGEBRA. 


OP   AlimniETICAL   I'KOrOKTION   AXD   PIIOGKESSION. 


If  we  have  four  numbers,  2,  4,  8,  and  10,  of  which 
the  difference  betM'een  the  first  and  second  is  equal  to 
the  difference  between  the  third  and  fourth,  these  numbers 
are  said  to  be  in  arithmetical  proportion.  The  first  term  2 
is  called  an  antecedent,  and  the  second  term  4,  with  which 
it  is  compared,  a  consequent.  The  number  8  is  also  called 
an  antecedent,  and  the  number  10,  with  which  it  is  com- 
pared, a  consequent. 

When  the  difference  between  the  first  and  second  is  equal 
to  the  difference  between  the  third  and  fourth,  the  four 
mimbers  are  said  to  be  in  proportion.    Thus,  the  numbers, 
2,     4,     8,     10, 

are  in  arithmetical  proportion. 

178.  When  the  difference  between  the  first  antecedent 
and  consequent  is  the  same  as  between  any  two  consecutive 
terms  of  the  proportion,  the  proportion  is  called  an  arith- 
metical progression.  Hence,  a  progression  by  differences, 
or  an  arithmetical  %>rogression,  is  a  series  in  which  the  suc- 
cessive terms  are  continually  increased  or  decreased  by  a 
constant  number,  which  is  called  the  common  difference  of 
the  progression. 

Thus,  in  the  two  series, 

1,       4,       7,      10,     13,     16,     19,     22,     25,  ... 
60,     56,     52,     48,     44,     40,     36,     32,     28,  ... 

177.  When  are  four  numbers  in  arithmetical  proportion  ?     What  is  the 
first  called?     What  is  the  second  called?     What  is  the  third  called? 
What  is  the  fourth  called  ? 

178.  What  is  an  arithmetical  progression  ?     What  is  the  number  called 
by  which  the  terms  are  increased  or  diminished  ?     What  is  an  increasing 
progression?     What  is  a  decreasing  progression?     Which  term  is  only 
an  antecedent  ?     Which  only  a  consequent  ? 


A  K  I  T  II  M  K  T  I  C  A  L       P  It  O  G  K  K  S  S  I  O  N.  26 1 

the  first  is  called  an  increasing  progression,  of  which  the 
common  difference  is  3,  and  the  second,  a  decreasing  pro- 
gression, of  which  the  common  difference  is  4. 

In  general,  let  a,  #,  c,  d,  e,  f,  ...  denote  the  terms  of 
a  progression  by  differences ;  it  has  been  agreed  to  write 
them  thus : 

a  .  b  .  c  .  d  .  e  .  f  .  g  .  h  .  i  .  k  .  .  . 

This  series  is  read,  a  is  to  b,  as  b  is  to  c,  as  c  is  to  d,  as  d 
is  to  e,  &c.  This  is  a  series  of  continued  equi-differences,  in 
which  each  term  is  at  the  same  time  an  antecedent  and  a 
consequent,  with  the  exception  of  the  first  term,  which  is 
only  an  antecedent,  and  the  last,  which  is  only  a  consequent. 

179.  Let  d  denote  the  common  difference  of  the .  pro- 
gresion, 

a.b.c.e.f.g.h.  &c., 

which  we  will  consider  increasing. 

From  the  definition  of  the  progression,  it  evidently  fol 
lows  that, 

b  =  a  +  d,    c  =  b  +  d  —  a  -f  2d,    e  =  c  +  d  =  a  +  3d; 

and,  in  general,  any  term  of  the  series  is  equal  to  the  first 
term,  plus  as  many  times  the  common  difference  as  there  are 
preceding  terms. 

Thus,  let  I  be  any  term,  and  n  the  number  which  marks 
the  place  of  it ;  the  expression  for  this  general  term  is, 

I  =  a  +  (n  —  l)d. 
Hence,  for  finding  the  last  term,  we  have  the  following 

179.  Give  the  rule  for  finding  the  last  term  of  a  series  when  the  pro- 
gression is  increasing. 


262  ELEMENTARY      ALGEBRA. 


KULE. 

I.   Multiply  the  common  difference  by  the  number  of 
terms  less  one: 

IT.    To  the  product  acid  the  first  term  y  the  sum  will  be 
the  last  term. 

EXAMPLES. 

The  formula, 

I  =  a  +  (n  —  !)<£, 

serves  to  find  any  term  whatever,  without  determining  all 
those  which  precede  it. 

1.  If  we  make    n  =  1,    we  have,    I  =  a ;    that  is,  the 
series  will  have  but  one  term. 

2.  If  we  make    n  =  2,    we  have,    I  =  a  +  d ;  that  is, 
the  series  will  have  two  terms,  and  the  second  term  is  equal 
to  the  first,  plus  the  common  difference. 

3.  If   a  =  3,    and    d  —  2,    what  is  the  3d  term? 

Ans.    7, 

4.  If   a  =  5,    and    d  =  4,    what  is  the  6th  term? 

Ans.    25. 

5.  If  a  —  7,    and    d  =  5,    what  is  the  9th  term  ? 

Ans.   47. 

6.  If  a  =  8,  and    d  =  5,    what  is  the  10th  term  ? 

Ans.    53. 

7.  If   a  =  20,    and    d  —  4,    what   is  the  12th  term? 

Ans.   64. 

8.  If   a  =  40,    and    d  =  20,    what  is  the  50th  term  ? 

Ans.    1020. 

0    If   a  =  45,    and    d  =  30,    what  is  the  40th  term? 

Ans.    1215. 


ARITHMETICAL      PROGRESSION.  263 

10.  If   a  =  30,    and    d  =  20,    what  is  the  60th  term? 

Ans.    1210. 

11.  If   a  —  50,    and    d  =  10,    what  is  the  100th  term? 

Ans.   1040. 

12.  To  find  the  50th  term  of  the  progression, 

1   .  4  .  7  .  10  .  13  .  16  .  19  .  .  . 
<ve  have,  I  =  1+49x3  =  148. 

13.  To  find  the  60th  term  of  the  progression, 
1  .  5  .  9  .  13  .  17  .  21  .  25  .  .  . 
we  have,  ^=1  +  59x4  =  237. 

18O.  If  the  progression  were  a  decreasing  one,  we 
should  have, 

I  =  a  —  (n  —  l)d. 

Hence,  to  find  the  last  term  of  a  decreasing  progression,  we 
have  the  following 

BULE. 

I.  Multiply  the  common  difference  by  the  number  of  terms 
less  one : 

IT.  Subtract  the  product  from  tJie  first  term ;  ike  re- 
mainder will  be  the  last  term. 

EXAMPLES. 

1.  The  first  term  of  a  decreasing  progression  is  60,  the 
number  of  terms  20,  and  the  common  difference  3  :  what  is 
the  last  term  ? 

l  =  a—(n  —  i)di   gives  1=  60  —  (20  —  1)3  =  60  -57  =  3. 

180.  Give  the  rule  for  finding  the  last  term  of  a  series,  when  the  pro- 
gression is  decreasing. 


264  ELEMENTARY       ALGEBRA. 

2.  The  first  term  is  90,  the  common  difference  4,  and  the 
number  of  terms  15  :  what  is  the  last  term  ?  Ans.   34. 

3.  The  first  term  is  100,  the  number  of  terms  40,  and  the 
common  difference  2 :  what  is  the  last  term  ?          Ans.   22. 

4.  The  first  term  is  80,  the  number  of  terms  10,  and  the 
common  difference  4  :  what  is  the  last  term  ?          Ans.   44. 

5.  The  first  term  is  600,  the  number  of  terms  100,  and 
the  common  difference  5  :  what  is  the  last  term  ? 

Ans.    105. 

6.  The  first  term  is  800,  the  number  of  terms  200,  and 
the  common  difference  2  :  what  is  the  last  term  ? 

Ans.   402. 

181.  A  progression  by  differences  being  given,  it  is  pro- 
posed to  prove  that,  the  sum  of  any  two  terms,  taken  at 
equal  distances  from  the  tico  extremes,  is  equal  to  the  sum 
of  the  two  extremes. 

That  is,  if  we  have  the  progression, 

2  .  4  .  6  .  8  .  10  .  12, 
we  wish  to  prove  generally,  that, 

4+10,     or     6  +  8, 
is  equal  to  the  sum  of  the  two  extremes,  2  and  12. 

Let  a.b.c.e.f...  i  .  k  .  /,  be  the  proposed 
progression,  and  n  the  number  of  terms. 

"We  will  first  observe  that,  if  x  denotes  a  term  which  has 
p  terms  before  it,  and  y  a  term  which  has  p  terms  after  it, 
we  have,  from  what  has  been  said, 

181.  In  every  progression  by  differences,  what  is  the  sum  of  the  two 
extremes  equal  to  ?  What  is  the  rule  for  finding  the  sum  of  an  arith 
metical  series? 


ARITHMETICAL      PRtGRESSION.  265 

X    =    a  +  p   X   <?, 

and,  y  =  I  —  p  X  c?; 

whence,  by  addition,     a;  -f  y  =  a  -f  J, 
which  proves  the  proposition. 

Referring  to  the  previous  example,  if  we  suppose,  in  the 
first  place,  x  to  denote  the  second  term  4,  then  y  will  de- 
note the  term  10,  next  to  the  last.  If  x  denotes  the  third 
term  6,  then  y  will  denote  8,  the  third  term  from  the  last. 

To  apply  this  principle  in  finding  the  sum  of  the  terms 
of  a  progression,  write  the  terms,  as  below,  and  then 
again,  in  an  inverse  order,  viz.  : 

a  .  b  .  c  .  d  .  e  ./...*.  k  .  /. 
/    .  Jc  .  i    .........  c  .  b  .  a. 

Calling  S  the  sum  of  the  terms  of  the  first  progression, 
2S  will  be  the  sum  of  the  terms  of  both  progressions,  and 
we  shall  have, 


Now,  since  all  the  parts,  a  +  ?,  b  +  k,  c  +  i  .  .  .  are 
equal  to  each  other,  and  their  number  equal  to  ?z, 

2S  =  (a  +  1)  X  n,     or     S  =  (—  j-J  x  n- 

Hence,  for  finding  the  sum  of  an  arithmetical  series,  we 
have  the  following 

KULE. 

I.  Add  the  two  extremes  together,  and  take  half  their  sum  : 

II.  Multiply  this  half-sum  by  the  number  of  terms  ;  tJie 
product  imtt  be  the  sum  of  the  strip*. 
12 


ELEMENTARY       ALGEBRA. 


i:  X  A  M  P  L  E  S  . 

1.  The  extremes  are  2  and  16,  and  the  number  of  terms 
8  :  what  is  the  sum  of  the  series  ? 

I  Ct       l~     '  \  •  **         I         J-  v 

;ives     S  —  — — —  X  8   =  72. 

2i 

2.  The  extremes  are,  3  and  27,  and  the  number  of  terms 
12  :  what  is  the  sum  of  the  series  ?  Ans.    180. 

3.  The  extremes  are  4  and  20,  and  the  number  of  terms 
10:  what  is  the  sum  of  the  series?  Ans.    120. 

4.  The  extremes  arc   100  and  200,  and  the  number  of 
terms  80:  what  is  the  sum  of  the  series?  Ans.    12000. 

5.  The  extremes  are  500  and  60,  and  the  number  of  terras 
20  :  what  is  the  sum  of  the  series  ?  Ans.   5600 

6.  The  extremes  are  800  and  1200,  and  the  number  of 
terms  50 :  what  is  the  sum  of  the  series?          Ans.   50000. 

1§2.  In  arithmetical  proportion  there  are  five  members 
to  be  considered : 

1st.   The  first  term,  «. 

2d.    The  common  difference,  d. 

3d.    The  number  of  terms,  n. 

4th.  The  last  term,  /. 

5th.  The  sum,  S. 

The  formulas, 

I  =  a  +  (n  —  l)d,    and     S  =  {-     —  J  x  n, 

contain  five  quantities,  o,  d,  ??,  /,  and  /S>,  and  consequently 
give  rise  to  the  following  general  problem,  viz. :  Any  three 

182.  How  many  numbers  are  considered  in  arithmetical  proportion? 
What  are  they  ?  In  every  arithmetical  progression,  \vhat  is  the  common 
difference  equal  to  ? 


ARITHMETICAL      PROGRESSION.  2(17 

of  these  Jive  quantities  briny  given,  to  determine  the  other 
tico. 

We  already  know  the  value  of  S  in  terms  of  a,  n,  and  I. 
From  the  formula, 

I  =  a  +  (n  —  I)c7, 
we  find,  a  =    I  —  (n  —  \}d. 

That  is :  The  first  term  of  an  increasing  arithmetical  pro- 
gression is  equal  to  the  last  term,  minus  the  product  of  the 
common  difference  by  the  number  of  terms  less  one. 

From  the  same  formula,  we  also  find, 

I  —  a 


d  = 


n  —  1 


That  is  :  In  any  arithmetical  progression,  the  common  dif- 
ference is  equal  to  the  last  term,  minus  the  first  term,  divided 
by  the  number  of  terms  less  one. 

The  last  term  is  10,  the  first  term  4,  and  the  number  of 
terms  5 :  what  is  the  common  difference  ? 

The  formula, 


n  —  1 

16-4 
gives,  d  =  -  -  —  3. 

4 

2.  The  last  term  is  22,  the  first  term  4,  and  the  number 
of  terms  10  :  what  is  the  common  difference?  Ans.  2. 

183.  The  last  principle  affords  a  solution  to  the  follow- 
ing question : 

To  find  a  number  m  of  arithmetical  means  between  two 
given  numbers  a  and  b. 

183.  How  do  you  find  any  number  of  arithmetical  means  between  two 
given  numbers  ? 


268  E  1    K  M  K  X  T  A  K  Y       A  L  G  K  B  R  A  . 

To  resolve  this  question,  it  is  first  necessary  to  find  the 
common  difference.  Now,  we  may  regard  -a  as  the  first 
term  of  an  arithmetical  progression,  b  as  the  last  term,  and 
the  required  means  as  intermediate  terms.  The  number  of 
terms  of  this  progression  will  be  expressed  by  m  +  2. 

Now,  by  substituting  in  the  above  formula,  b  for  ?,  and 
m  +  2  for  n,  it  becomes, 

,  b  —  a  b  —  « 

m  +  2  —  1   '  ~  m  +  1  ' 

that  is  :  The  common  difference  of  the  required  progression 
is  obtained  by  dividing  the  difference  betioeen  the  given 
numbers,  a  and  b,  by  the  required  number  of  means  2>hts  one. 

Having  obtained  the  common  difference,  d,  form  the  second 
term  of  the  progression,  or  the  first  arithmetical  mean,  by 
adding  d  to  the  first  term  a.  The  second  mean  is  obtained 
by  augmenting  the  first  mean  by  d,  &c. 

1.  Find  three  arithmetical  means  between  the  extremes 
2  and  18. 

The  formula,  d  =   -  . 

m  +  1 

18-2 
gives,  £  =  —  —    =  4  ; 

hence,  the  progression  is, 

2  .  6  .  10  .  14  .  18. 

2.  Find  twelve  arithmetical  means  between  12  and  77. 


The  formula,  d  = 


m 


77  -  12 
gives,  d  =  --  -  —    =  5  ; 

hence,  the  progression  is, 

12  .  17  .  22  .  27  .         .  77. 


A  R  I  T  H  M  K  T  I  C  A  L       1'  K  O  G  R  K  S  S  I  O  N .  209 

184.  REMARK. — If  the  same  number  of  arithmetical 
means  are  inserted  between  all  the  terms,  taken  two  and 
two,  these  terms,  and  the  arithmetical  means  united,  will 
form  one  and  the  same  progression. 

For,  let  a  .  b  .  c  .  e  .  f .  .  .  be  the  proposed  progression, 
and  m  the  number  of  means  to  be  inserted  between  a  and 
£>,  b  and  c,  c  and  e  .  .  .  .  &c. 

From  what  has  just  been  said,  the  common  difference  of 
each  partial  progression  Avill  be  expressed  by 

b  —  a      c  —  b       e  —  c 
m  +  1 '     m  +  1 '     m  +  1  ' 

expressions  which  are  equal  to  each  other,  since  «,  5,  c  .  .  . 
are  in  progression  ;  therefore,  the  common  difference  is  the 
same  in  each  of  the  partial  progressions  ;  and,  since  the  last 
term  of  the  first  forms  the  first  term  of  the  second,  &c.,  we 
may  conclude,  that  all  of  these  partial  progressions  form  a 
single  progression. 

EXAMPLES. 

1.  Find  the  sum  of  the  first  fifty  terms  of  the  progression 
2  .  9  .  16  .  23  ... 

For  the  50th  term,  we  have, 

I  =  2  +  49  X  7   =   345. 

50 
Hence,     8  =   (2  +  345)  x  --  =  347  X  25  =  8675. 

£ 

2.  Find  the  100th  term  of  the  series  2  .  9  .  16  .  23  .... 

Ans.   695. 

3.  Find  the  sum  of  100  terms  of  the  series  1.3.5.7. 
9  .  Ans.   10000. 


270  K  L  ]-:  M  K  X  T  A  K  Y       ALGEBRA. 

4.  The  greatest  term  is  70,  the  common  difference  3,  nnd 
the  number  of  terms  21  :  what  is  the  least  term  and  the 
sum  of  the  series  ? 

Ans.   Least  term,  10  ;  sum  of  series,  840. 

5.  The  first  term  is  4,  the  common  difference  8,  and  the 
number  of  terms  8  :  what  is  the  last  term,  and  the  sum  of 
the  series?  Ans.  j  Last  term,    GO. 

(  Sum       =  25G. 

6.  The  first  term  is  2,  the  last  term  20,  and  the  number 
of  terms  10 :  what  is  the  common  difference  ?  Ans.    2. 

7.  Insert  four  means  between  the  two  numbers  4  and  1 9 : 
what  is  the  series?  Ans.   4  .  7  .  10  .  13  .  16  .  19. 

8.  The  first  term  of  a  decreasing  arithmetical  progression 
is  10,  .the  common  difference  one-third,  and  the  number  of 
terms  21 :  required  the  sum  of  the  series.  Ans.    140. 

9.  In  a  progression  by  differences,  having  given  the  com- 
mon difference  6,  the  last  term  185,  and  the  sum  of  the 
terms  2945  :  find  the  first  term,  and  the  number  of  terms. 

Ans.   First  term  =  5  ;  number  of  terms,  31. 

10.  Find  nine  arithmetical  means  between  each  antecedent 
and  consequent  of  the.  progression  2. 5. 8. 11.  14... 

Ans.    Common  diff.,  or  d  =  0.3. 

11.  Find  the  number  of  men  contained  in  a  triangular 
battalion,  the  first  rank  containing  one  man,  the  second  2, 
the  third  3,  and  so  on  to  the  nth,  which  contains  n.    In  other 
words,  find  the  expression  for  the  sum  of  the  natural  num- 
bers 1,  2,  3  .  .  .,  from  1  to  n  inclusively. 

Ans.    S  =  *±-L>. 

2 

12.  Find  the  sum  of  the  n  first  terms  of  the  progression 
of  uneven  numbers,  1.3.5.7.9,...          Ans.    8  =  ;i2. 


GEOMETRICAL     PROFOKTION.  271 

13.  One  hundred  stones  being  placed  on  the  ground  in  a 
straight  line,  at  the  distance  of  2  yards  apart,  how  far  will 
a  person  travel  who  shall  bring' them  one  by  one  to  a  basket, 
placed  at  a  distance  of  2  yards  from  the  first  stone  ? 

Ans.   11  miles,  840  yards. 


GEOMETRICAL     PROPORTION    AND     PROGRESSION. 

185.  jRatio  is  the  quotient  arising  from  dividing  one 
quantity  by  another  quantity  of  the  same  kind,  regarded 
as  a  standard.  Thus,  if  the  numbers  3  and  6  have  the  same 
unit,  the  ratio  of  3  to  6  will  be  expressed  by 


And  in  general,  if  A  and  It  represent  quantities  of  the  same 
kind,  the  ratio  of  A  to  J5  will  be  expressed  by 

B 
A' 

1§6.     The  character    cc   indicates  that  one  quantity  is 
proportional  to  another.     Thus, 

A  cc  .B, 
is  read,  A  proportional  to  JB. 

If  there  be  four  numbers, 

2,     4,     8,     16, 

having  such  values  that  the  second  divided  by  the  first  is 
equal  to  the  fourth  divided  by  the  third,  the  numbers  are 

185.  What  is  ratio  ?     What  is  the  ratio  of  3  to  G  ?     Of  4  to  12  ? 

186.  What  is  proportion?     How  do  you  express  that  four  numbers 
are  in  proportion  ?     What  are  the  numbers  callnd?     What  are  the  first 
•ud  fourth  terms  cilled  ?    What  the  second  and  third  ? 


272  ELEMENTAKY      ALGEBRA 

said  to  form  a  proportion.  And  in  general,  if  there  be  four 
quantities,  A,  J?,  <7,  and  .Z),  having  such  values  that, 

B        D 

A  '"''   <?' 

then,  A  is  said  to  have  the  same  ratio  to  B  that  C  has  to  D\ 
or,  the  ratio  of  A  to  B  is  equal  to  the  ratio  of  C  to  D. 
When  four  quantities  have  this  relation  to  each  other,  com- 
pared  together  two  and  two,  they  are  said  to  form  a  geo- 
metrical proportion. 

To  express  that  the  ratio  of  A  to  B  is  equal  to  the  ratio 
of  C  to  _D,  we  write  the  quantities  thus, 

A  :  B  ::    C  :  D; 
and  read,  A  is  to  B  as  C  to  D. 

The  quantities  which  are  compared,  the  one  with  the 
other,  are  called  terms  of  the  proportion.  The  first  and  last 
terms  are  called  the  two  extremes,  and  the  second  and  third 
terms,  the  two  means.  Thus,  A  and  D  are  the  extremes, 
and  B  and  C  the  means. 

187.  Of  four  terms  of  a  proportion,  the  first  and  third 
are  called  the  antecedents,  and  the  second  and  fourth  the 
consequents  ;  and  the  last  is  said  to  be  a  fourth  proportional 
to  the  other  three,  taken  in  order.     Thus,  in  the  last  pro- 
portion A  and  C  are  the  antecedents,  and  B  and  D  the  con- 
sequents. 

188.  Three  quantities  are  in  proportion,  when  the  first 
has  the  same  ratio  to  the  second  that  the  second  has  to  the 

"187.  In  four  proportional  quantities,  what  are  the  first  and  third  called? 
What  the  second  and  fourth  ? 

188.  When  are  three  quantities  proportional?  What  is  the  middle  one 
called  ? 


GEOMETRICAL      PROPORTION.  273 

third ;  and  then  the  middle  term  is  said1  to  be  a  mean  pro- 
portional between  the  other  two.     For  example, 

3  :  6  :  :  6  :  12; 
and  6  is  a  mean  proportional  between  3  and  12.  - 

189.  Four  quantities  are  said  to  be  in  proportion  by  in- 
version,  or  inversely,  when  the  consequents  are  made  the 
antecedents,  and  the  antecedents  the  consequents. 

Thus,  if  we  have  the  proportion, 

3  :  6  :  :  8  :  16, 
the  inverse  proportion  would  be, 

6  :  3  :  :  1C  :  8. 

190.  Quantities  are  said  to  be  in  proportion  by  alterna- 
tion, or  alternately,  when  antecedent  is  compared  with  ante- 
cedent, and  consequent  with  consequent. 

Thus,  if  we  have  the  proportion, 

3  :  6  :  :  8  :  16, 
the  alternate  proportion  would  be, 

3  :  8   :  :  6  :   16. 

191.  Quantities  are  said  to  be  in  proportion  by  compo- 
sition, when  the  sum  of  the  antecedent  and  consequent  is 
compared  either  with  antecedent  or  consequent 

Thus,  if  we  have  the  proportion, 

2  :  4  :  :  8  :  16, 

189.  When  are  quantities  said  to  be  in  proportion  by  inversion,  or  in 
versely  ? 

190.  When  are  quantities  in  proportion  by  alternation? 

191.  When  are  quantities  in  proportion  by  composition? 

12* 


ELEMENTAKY      ALGEBRA. 

the  proportion  by  composition  would  be, 

2  +  4  :  2  :  :  8  +  16  :  8; 
and,  2  +  4  :  4  :  :  8  +  16  :  16. 

192.  Quantities  are  said  to  be  in  proportion  by  division, 
when  the  difference  of  the  antecedent  and  consequent  is 
compared  either  with  antecedent  or  consequent. 

Thus,  if  we  have  the  proportion, 

3  :  9  :  :  12  :  36, 
the  proportion  by  division  will  be, 

9  —  3  :  3  :  :  36  —  12  :  12; 
and,  9  —  3  :  9  :  :  36  —  12  :  36; 

193.  Equi-multiples  of  two  or  more  quantities  are  the 
products  which  arise  from  multiplying  the  qiiantities  by  the 
same  number. 

Thus,  if  we  have  any  two  numbers,  as  6  and  5,  and  mul- 
tiply them  both  by  any  number,  as  9,  the  equi-multiples  will 
be  54  and  45 ;  for, 

6  X  9   =   54,    and    5  X  9   =   45. 

Also,    m  x  A,    and    m  x  JB,    are  equi-multiples  of  A  and 
1?,  the  common  multiplier  being  m. 

• 

194.  Two  quantities  A  and  J?,  which  may  change  their 
values,  are  reciprocally  or  inversely  proportional,  when  one 
is  proportionaljo  unity  divided  by  the  other,  and  then  their 
product  remains  constant. 

192.  When  are  quantities  in  proportion  by  division  ? 

193.  What  arc  equi-multiples  of  two  or  more  quantities? 

194.  When  are  two  quantities  said  to  be  reciprocally  proportional? 


GEOMETRICAL      PROPORTION.  275 

"We  express  this  reciprocal  or  inverse  relation  thus, 

A  ocl. 
in  which  A  is  said  to  be  inversely  proportional  to  1$. 

195.  If  we  have  the  proportion, 

A  :  B  :  :   C  :  D, 

T)  J\ 

we  have,  f  =  -^,    (Art.  186); 

•A.  C 

and  by  clearing  the  equation  of  fractions,  we  have, 
BG  =  AD. 

That  is  :  Of  four  proportional  quantities,  the  product  of 
the  two  extremes  is  equal  to  the  product  of  the  two  means. 

Tliis  general  principle  is  apparent  in  the  proportion  be- 
tween the  numbers, 

2  :  10  :  :  12  :  60, 
which  gives,       2  x  60  —   10  X  12  =  120. 

196.  If  four  quantities,  A,  B,  (7,  D,  are  so  related  to 
each  other,  that 

A  x  D  =  B  x  C, 

we  shall  also  have, 

A        C 

and  hence,  A  :  B  :  :   C  :  D. 

That  is :  If  the  product  of  two  quantities  is  equal  to  the 
product  of  two  other  quantities,  two  of  them  may  be  made 
the  extremes,  and  the  other  tico  the  means  of  a  proportion. 

195.  If  four  quantities  are  proportional,  what  is  the  product  of  the  two 
means  equal  to  ? 

196.  If  the  product  of  two  quantities  is  equal  to  the  product  of  two 
^ther  quantities,  may  the  four  be  placed  in  a  proportion  ?     How  ? 


276  ELEMENTARY      ALGEBRA. 

Thus,  if  we  have, 

2x8  =  4x4, 
we  also  have, 

2  :  4  :  :  4  :  8. 

197.  If  we  have  three  proportional  quantities, 

A  :  B  :  :  B  :   C, 

B         C 

we  have,  -j  =  -^ ; 

A         J$ 

hence,  Bz  =  AC. 

That  is:   If  three  quantities  are  proportional,  the  square  of 
the  middle  term  is  equal  to  the  product  of  tJie  two  extremes. 

Thus,  if  we  have  the  proportion, 

3  :  6  :  :  6  :  12, 
we  shall  also  have, 

6  X  6   =  62  =   3  X  12   =  36. 

198.  If  we  Lave, 

7?  Tl 

A  :  B  :  :   C  :  D,    and  consequently,    :  T  =  -^ , 

.4         C 

multiply  both  members  of  the  last  equation  by    -= ,    and 

we  then  obtain, 

CD 
A  "  B' 

and,  hence,  A  :  C  : :  B  :  D. 

That  is  :  If  four  quantities  are  proportional,  they  icitt  be 
in  proportion  by  alternation. 

197.  If  three  quantities  are  proportional,  what  is  the  product  of  the 
extremes  equal  to  ? 

198.  If  four  quantities  are  proportional,  will  they  oe  in  proportion  by 
alternation  ? 


GEOMETRICAL      PROPORTION.  277 

Let  us  take,  as  an  example, 

10  :  15  : :  20  :  30. 

We  shall  have,  by  alternating  the  terms, 
10  :  20  :  :  15  :  30. 

199.  If  we  have, 

A  :  B  :  :  C  :  D,    and    A  :  B  :  :  E :  F, 
we  shall  also  have, 

B        D  B        F 

—  =  -^ ,    and    -  T  =  -= ; 
A         C  A        E 

~T\  77T 

hence,  -^  =  — ,    and    C  :  D  :  :  E  :  F. 

L>          xL 

That  is :  If  there  are  two  sets  of  proportions  having  an  an- 
tecedent and  consequent  in  the  one,  equal  to  an  antecedent 
and  consequent  of  the  other,  the  remaining  terms  will  be 
proportional. 
If  we  have  the  two  proportions, 

2  :  6  :  :  8  :  24,    and    2  :  6  :  :  10  :  30, 

we  shall  also  have, 

8  :  24  :  :  10  :  30. 

200.  If  we  have, 

7?  J\ 

A  '.  B  :  :   C  :  D,    and    consequently,    -j  =  -^, 

we  have,  by  dividing  1  by  each  member  of  the  equation, 

A         C 

-=  =    _  ,    and  consequently,    B  :  A  :  :  D  :   C. 

199.  If  you  have  two  sets  of  proportions  having  an  antecedent  and  con- 
sequent in  each,  equal ;  what  will  follow  ? 

200.  If  four  quantities  are  in  proportion,  will  they  be  in  proportion 
when  taken  inversely  ? 


278  ELEMENTARY       ALGEBRA. 

That  is :  Four  proportional  quantities  will  b$  in  proportion^ 
when  taken  inversely. 

To  give  an  example  iu  numbers,  take  the  proportion, 

7  :  14  :  :  8  :  16; 
then,  the  inverse  proportion  will  be, 

14  .  7  :  :  16  :  8, 
hi  which  the  ratio  is  one-half. 

2O1.     The  proportion, 

A  :  B  :  :   C  :  D,    gives,    A  x  D  =  B  x  C. 

To  each  member  of  the  last  equation  add  H  x  D.    "We 
shall  then  have, 

(A  +  J?)  x  D  =  (C  +  D)  x  JB; 
and  by  separating  the  factors,  we  obtain, 

A  +  B  :  B  :  :   C  +  D  :  D. 

If,  instead  of  adding,  we  subtract  JB  x  D  from  both 
members,  we  have, 

(A  -  B)  x  D  =  (C-  D)  x  J?; 
which  gives, 

A  -  B  :  JB  :  :   C  -  D  :  Z>. 

That  is:  If  four  quantities  are  proportional,  they  will  be 
in  proportion  by  composition  or  division. 

Thus,  if  we  have  the  proportion, 

9  :  27  :  :  16  :  48, 

21'!.  If  four  quantities  are  in  proportion,  will  they  be  in  proportion  by 
composition  ?  Will  they  be  in  proportion  by  division  ?  What  is  th« 
difl'ereiice  between  composition  and  division  ? 


GEOMETRICAL      PROPORTION.  279 

tre  shall  have,  by  composition, 

9  +  27  :  27  :  :  16  +  48  :  48; 
that  .is,  36  :  27  :  :  64  :  48, 

in  which  the  ratio  is  three-fourths. 
The  same  proportion  gives  us,  by  division, 

27  —  9  :  27  ::  48  —  16  :  48; 
that  is,  18  :  27  :  :  32  :  48, 

in  which  the  ratio  is  one  and  one-half. 
2O2.     If  we  have, 


and  multiply  the  numerator  and  denominator  of  the  first 
member  by  any  number  m,  we  obtain, 

j   =  -=  ,    and    mA  :  mJ3  :  :   C  :  D. 
mA         C 

That  is  :  Equal  multiples  of  two  quantities  have  the  same 
ratio  as  the  quantities  themselves. 

For  example,  if  we  have  the  proportion, 
5  :  10  :  :  12  :  24, 

and  multiply  the  first  antecedent  and  consequent  by  6,  we 
have, 

30  :  60  :  :  12  :  24, 

in  which  the  ratio  is  still  2. 

2O3.     The  proportions, 

A  :  B  :  :   C  :  D,    and    A  :  B  :  :  E  :  F, 

202.  Have  equal  multiples  of  two  quantities  the  same  ratio  as  the 
quantities  ? 

203  Suppose  the  antecedent  and  consequent  be  augmented  or  dimin- 
ished by  quantities  having  the  same  ratio  ? 


280  ELEMENTARY      ALGEBRA. 

give,     A  x  D  =  JB  x  <7,     and    AxF=JSxE\ 
adding  and  subtracting  these  equations,  we  obtain, 
A(D+F)  =  E(C±E],     or    A  :  B  :  :  G±  E  :  D±  F. 

That  is :  If  C  and  D,  the  antecedent  and  consequent,  be 
augmented  or  diminished  by  quantities  E  and  F,  which 
have  the  same  ratio  as  C  to  D,  the  resulting  quantities  will 
also  have  the  same  ratio. 

Let  us  take,  as  an  example,  the  proportion, 

9  :  18  : :  20  :  40, 
in  which  the  ratio  is  2. 

If  we  augment  the  antecedent  and  consequent  by  the 
numbers  15  and  30,  which  have  the  same  ratio,  we  shall 
have, 

9  -f  15  :  18  +  30  :  :  20  :  40; 

that  is,  24  :  48   :  :  20  :  40, 

in  which  the  ratio  is  still  2. 

If  we  diminish  the  second  antecedent  and  consequent  by 
these  numbers  respectively,  we  have, 

9  :  18  ::  20  —  15  :  40  —  30; 
that  is,  9  :  18  :  :  5  :  10, 

in  which  the  ratio  is  till  2. 

2O4.    If  we  have  several  proportions, 
A  :  B  :  :  C  :  D,    which  gives    A  x  D  —  B  x  (7, 
A  :  B  : :  E :  F,     which  gives    A  x  F  =  B  x  E, 
A  :  B  : :  G  :  H,    which  gives    A  x  H  =  B  x  G, 
&c.,  &c., 

204.  In  any  number  of  proportions  having  the  same  ratio,  how  viD 
any  one  antecedent  be  to  its  consequent? 


GEOMETRICAL       PROPORTION.  281 

we  shall  have,  by  addition, 

A(D  +  F+  II)  =  B(C  +  E  +  G); 
and  by  separating  the  factors, 

A  :  B  :  :   C  +  E  +  G  :  D  +  F  +  H. 

That  is:  In  any  number  of  proportions  having  the  same 
ratio,  any  antecedent  will  be  to  its  consequent  as  the  sum 
of  the  antecedents  to  the  sum  of  the  consequents. 

Let  us  take,  for  example, 

2  :  4  :  :  6  :  12,     and     1  :  2  :  :  3  :  6,    &c. 
Then  2:4::6  +  3:12  +  6; 

that  is,  2  :  4  :  :  9  :  18, 

in  which  the  ratio  is  still  2. 

2O5.     If  we  have  four  proportional  quantities, 
A  :  B  :  :   C  :  D,    we  have,      -.   —  -^ ; 

^3L  0 

and  raising  both  members  to  any  power  whose  exponent  is 
n,  or  extracting  any  root  whose  index  is  n,  we  have, 

j?»        D» 

-^  =-.  -£-,     and  consequently, 

An  :  Bn  :  :  Cn  :  Dn. 

That  is :  If  four  quantities  are  proportional,  their  like 
powers  or  roots  will  be  proportional. 

If  we  have,  for  example, 

2:4    :  :  3    :  6, 
we  shall  have,  22  :  42  :  :  32  :  62 ; 

205.  In  four  proportional  quantities,  how  arc  like  powers  or  roots? 


ELEMENTARY       A  L  G  .E  B  H  A . 

that  is,  4   :  16  :  :  9    :  36, 

in  which  the  terms  are  proportional,  the  ratio  being  4. 

2O6.     Let  there  be  two  sets  of  proportions, 

7?          T) 

A  :  J5  :  :   C  :  D,    which  gives       r  =  -~ ; 

jfl          0 

F       II 

E  :  F  :  :  G  :  JET,     which  gives      -=,  =   -^  • 

.&        Or 

Multiply  them  together,  member  by  member,  we  have, 

E  x  F  _    D  x  H 
A  x  E  z:   G  x~£' 

A  x  E  :  B  x  F::   C  x  G  :  D  x  H. 

Tliat  is  :  In  two  sets  of  proportional  quantities,  the  product* 
of  the  corresponding  terms  are  proportional. 

Thus,  if  we  hav^  the  two  proportions, 
8  :  16  :  :  10  :     20, 

and,  3  :     4  :  :     6  :       8, 

we  shall  have,         24  :  64  :  :  60  :  160. 


GEOMETRICAL     PROGRESSION. 

2O71.  We  have  thus  far  only  considered  the  case  in  which 
the  ratio  of  the  first  term  to  the  second  is  the  same  as  that 
of  the  third  to  the  fourth. 

20C.  In  two  sets  of  proportions,  how  arc  the  products  of  the  correspond- 
ing terms  ? 

207.  What  is  a  geometrical  progression?  What  is  the  ratio  of  the 
progression  ?  If  any  term  of  a  progression  be  multiplied  by  the  ratio, 
what  will  the  product  be  ?  If  any  term  bo  divided  by  the  ratio,  w  hat 


G  E  U  M  K  T  K  I  C  A  L      I'  K  O  G  B  K  6  S  I  O  N.  283 

If  we  1m  e  the  farther  condition,  that  the  ratio  of  the 
second  term  to  the  third  shall  also  be  the  same  as  that  of 
the  first  to  the  second,  or  of  the  third  to  the  fourth,  we  shall 
have  a  series  of  numbers,  each  one  of  which,  divided  by 
the  preceding  one,  will  give  the  same  ratio.  Hence,  if  any 
term  be  multiplied  by  this  quotient,  the  product  will  be  the 
succeeding  term.  A  series  of  numbers  so  formed,  is  called 
a  geometrical  progression.  Hence, 

A  Geometrical  Progression,  or  progression,  by  quotients, 
is  a  series  of  terms,  each  of  which  is  equal  to  the  preceding 
term  multiplied  by  a  constant  number,  which  number  is 
called  the  ratio  of  the  progression.  Thus, 

1  :  3  :  9  :  27  :  81  :  243,  &c., 

is  a  geometrical  progression,  in  which  the  ratio  is  3.  It  is 
written  by  merely  placing  two  dots  between  the  terms. 

Also,  64  :  32  :  16  :  8  :  4  :  2  :  1, 

is  u  geometrical  progression  in  which  the  ratio  is  one-half. 

In  the  first  progression  each  term  is  contained  three  times 
in  the  one  that  follows,  and  hence  the  ratio  is  3.  In  the 
second,  each  term  is  contained  one-half  times  in  the  one 
which  follows,  and  hence  the  ratio  is  one-half. 

The  first  is  called  an  increasing  progression,  and  the 
second  a  decreasing  progression. 

Let  a,  b,  c,  d,  e,  f,  ...  be  numbers,  in  a  progression  by 
quotients  ;  they  are  written  thus  : 

a  :  b  :  c  :  d  :  e  :  f  :  g  .  .  . 

and  it  is  enunciated  in  the  same  manner  as  a  progression  by 
differences.  It  is  necessary,  however,  to  make  the  distinc- 

will  the  quotient  be?  How  is  a  p-ogression  by  quotients  written?  Which 
of  the  terms  is  only  nu  antecedent?  Which  only  a  consequent?  Hovt 
laav  each  of  the  others  be  considered? 


284  ELEMENTARY       ALGEBRA.. 

tion,  that  one  is  a  series  formed  by  equal  differences,  and 
the  other  a  series  formed  by  equal  quotients  or  ratios.  It 
should  be  remarked  that  each  term  is  at  the  same  time  an 
antecedent  and  a  consequent,  except  the  first,  which  is  only 
an  antecedent,  and  the  last,  which  is  only  a  consequent. 

2O8.     Let  r  denote  the  ratio  of  the  progression, 

a  :  b  :  c  :  d  .  .  . 

r  being  >  1  when  the  progression  is  increasing,  and  r<  1 
when  it  is  decreasing.  Then,  since, 

b  _  c  d  e  . 

a  b  c  d  ~ 

we  have, 

b  =  ar,    c  =  br  —  ar2,     d  =  cr  =  ar3,     e  =  dr  =  ar*, 
f  =  er  =  ar5  .  .  . 

that  is,  the  second  term  is  equal  to  ar,  the  third  to  ar2,  the 
fourth  to  ar3,  the  fifth  to  ar4,  &c. ;  and  in  general,  the  nth 
term,  that  is,  one  which  has  n  —  I  terms  before  it,  is  ex- 
pressed by  arn  ~ 1. 

Let  I  be  this  term  •  we  then  have  the  formula, 

I  =  arn~\ 

by  means  of  which  we  can  obtain  any  term  without  being 
obliged  to  find  all  the  terms  which  precede  it.  Hence,  to 
find  the  last  term  of  a  progression,  we  have  the  following 

E  u  L  E  . 

I.  Raise  the  ratio  to  a  power  whose  exponent  is  one  less 
than  the  number  of  terms. 

H.  Multiply  the  power  thus  found  by  the  first  term :  the 
product  will  be  the  required  term. 

208.  By  what  letter  do  we  denote  the  ratio  of  a  progression?  In  an 
increasing  progression  is  r  greater  or  less  than  1  ?  In  a  decreasing  pro- 


GEOMETRICAL      PROGRESSION.  285 

EXAMPLES. 

1.  Find  the  5th  term  of  the  progression, 

2  :  4  :  8  :  16  .  .  . 

in  which  the  first  term  is  2,  and  the  common  ratio  2. 
5th  term  =  2  X  2*  =  2  X  16  =  32.   Ans. 

2.  Find  the  8th  tetm  of  the  progression, 

2  :  6  :  18  :  54  ... 

8th  term  =  2  X  37  =  2  X  2187  =  4374.   Ans. 

3.  Find  the  6th  term  of  the  progression, 

2  :  8  :  32  :  128  ... 
6th  term  =  2  X  45  =  2  X  1024   =  2048.   Ant 

4.  Find  the  7th  term  of  the  progression, 

3  :  9  :  27  :  81  .  .  . 

7th  term  =  3  x  36  =  3  x  729  =  2187.   Ans. 

5.  Find  the  6th  term  of  the  progression, 

4  :  12  :  36  :  108  ... 
6th  term  =  4  X  3s  —  4  x  243   =  972.   Ans. 

6.  A  person  agreed  to  pay  his  servant  1  cent  for  the  first 
day,  two  for  the  second,  and  four  for  the  third,  doubling 
every  day  for  ten  days :  how  much  did  he  receive  on  the 
tenth  day?  Ans.   $5.12. 

grcssiou  is  r  greater  or  less  than  1  ?  If  a  is  the  first  term  and  r  the 
ratio,  what  is  the  second  term  equal  to  ?  What  the  third  ?  What  the 
fourth  ?  What  is  the  la?t  term  equal  to  ?  Give  the  rule  for  finding  the 
last  terra. 


286  ELK  MJ:>  ;TA  i:  Y     ALGEBRA. 

7.  What  is  the  8th  term  of  the  progression, 

9  :  36  :  144  :  576  .  .  . 
8th  term  =   9  X  47  =   9  x  16384   -   147456.   Ans. 

8.  Find  the  12th  term  of  the  progression, 

64  :  16  :  4  :  1  :  7  . 
4 

/l\u         43  1  1 

12th  term  —  64-1     =   —   =  -   =  —  —  -  •   Ans. 
\4/  411         48         60086 

2O9.     We  will  now  proceed  to  determine  the  sum  of  n 
terms  of  a  progression, 

a  :  b  :  c  :  d  :  e  :  f  :  .  .  .  :  i  :  Jc  :  /; 
I  denoting  the  nih  term. 

We  have  the  equations  (Art.  208), 
b  —  ar,     c  =  br,     d  =  cr,     e  —  dr,  .  .  .  k  =  ir,     I  =  AT, 

and  by  adding  them  all  together,  member  to  member,  we 
deduce, 

Sum  of  1st  members.  Sum  of  M  members. 

b+c+d+e+  .  .  .  +&+l=(a  +  b  +  c 


in  which  we  see  that  the  first  member  contains  all  the  terms 
but  «,  and  the  polynomial,  within  the  parenthesis  in  the 
second  member,  contains  all  the  terms  but  I.  Hence,  if  we 
call  the  sum  of  the  terms  S,  we  have, 

S  -  a  =  (S  -  l)r  =  Sr  -  Ir,      .-.  Sr  -  S  =  Ir  -  a; 

„        Ir  —  a 

whence,  o  =  --  • 

r  —  1 

209.  Give  the  rule  for  finding  the  sum  of  the  series.     What  is  the  first 
sttp?     What  the  second?     What  the  third  ? 


G  E  O  M  K  T  li  I  C  A  L      PROGRESSION.  287 

Therefore,  to  obtain  the  sum  of  all  the  terms,  or  sum  of  the 
series  of  a  geometrical  progression,  we  have  the 

KULE. 

I.  Multiply  the  last  term  by  the  ratio  : 
II.  Subtract  the  first  term  from  the  product : 
III.  Divide  the  remainder  by  the  ratio  diminished  by  1 
and  the  quotient  will  be  the  sum  of  the  series. 

1.  Find  the  sum  of  eight  terms  of  the  progression, 

2  :  6  :  18  :  54  :  1C2  .  .  .  2  x  37  =  4374. 

=  lr_-a  =   13IM-.   _  656fl_ 
'  r  —  1  2 

2.  Find  the  sum  of  the  progression, 

2  :  4  :  8  :  16  :  32. 

8=  lr~Cl  -  64~2   --   62 
'   r  -  1  1 

3.  Find  the  sum  of  ten  terms  of  the  progression, 

2  :  6  :  18  :  54  :  16^  .  .  .  2  X  39  =   39366. 

Ans.   59048. 

4.  What  debt  may  be  discharged  in  a  year,  or  twelve 
months,  by  paying  $1  the  first  month,  $2  the  second  month, 
$4  the  third  month,  and  so  on,  'each  succeeding  payment 
being  double  the  last ;  and  what  will  be  the  last  payment  ? 

i  Debt,      .  $4095. 

Ans.  *\  -. 

(  Last  payment,  $2048. 

5.  A  daughter  was  married  on  New-Year's  day.     Her 
father  gave  her  Is.,  writh  an  agreement  to  double  it  on  the 
first  of  the  next  month,  and  at  the  beginning  of  each  succeed- 
ing month  to  double  what  she  had  previously  received.    How 
much  did  she  receive  ?  Ans.   £204  15*. 


288  ELEMENTARY       ALGEBRA. 

6.  A  man  bought  ten  bushels  of  wheat,  on  the  condition 
that  he  should  pay  1  cent  for  the  first  bushel,  3  for  the  second, 
9  for  the  third,  and  so  on  to  the  last  :  what  did  he  pay  for 
the  last  bushel,  and  for  the  ten  bushels  ? 

j  Last  bushel,  $196  83. 
HS'  (  Total  cost,     $295  24. 

7.  A  man  plants  4  bushels  of  barley,  which,  at  the  first 
harvest,  produced  32  bushels  ;  these  he  also  plants,  which, 
in  like  manner,  produce  8  fold  ;  he  again  plants  all  his  crop, 
and  again  gets  8  fold,  and  so  on  for  16  years  :  what  is  his 
last  crop,  and  what  the  sum  of  the  series  ? 

.    s    j  Last,  140737488355328  bush. 
m'   (  Sum,  160842843834660. 

21O.     When  the  progression  is  decreasing,  we   have, 
r  <  1,  and  ?<  a  ;  the  above  formula, 

_  lr-g 

- 


for  the  sum,  is  then  written  under  the  form, 

a  —  Ir 

o  —  ^  -  1 
1  —  r 

in  order  that  the  two  terms  of  the  fraction  may  be  positive. 
1.  Find  the  sum  of  the  terms  of  the  progression, 
32  :  16  :  8  :  4  :  2 


32  —  2  X 
0        a  —  Ir  2         31 

S  =  -  -  =  --  -  --  zr  —  —  62. 
1  —  r                  1  1 


i      210.  What  is  the  formula  for  the  sum  of  the  series  of  a  decreasing 
'  progression  ? 


GEOMETRICAL       PROGRESSION.  289 

2.  Find  the  sum  of  the  first  twelve  terms  of  the  pro- 
gression, 


64  :  16  :  4  :  1  : 

4 


84/7  ),     or    —  l— 

\4/  65536 


X-       256-- 


S  —  a"~lr  _  65536 4  _  65536  _  65535 

=  1  -  r  ~  '~3~~  ~~  *"  196008 

4 

211.  REMARK. — We  perceive  that  the  principal  difficulty 
consists  in  obtaining  the  numerical  value  of  the  last  term,  a 
tedious  operation,  even  when  the  number  of  terms  is  not 
very  great. 

3.  Find  the  sum  of  six  terms  of  the  progression, 

512  :  128  :  32  ... 

AM.   682f 

4.  Find  the  sum  of  seven  terms  of  the  progression, 

2187  :  729  :  243  ... 

Ans.   3279. 

5.  Find  the  sum  of  six  tenns  of  the  progression, 

972  :  324  :  108  ... 

Ans.    1456. 

6.  Find  the  sum  of  eight  terms  of  the  progression, 

147456  :  36864  :  9216  .  .  . 

Ans.    196605. 

OF   PROGRESSIONS     HAVING   AN    INFINITE    NUMBER    OF  TERMS. 

212.  Let  there  be  the  decreasing  progression, 

a  :  b  :  o  :  d  :  e  :  f  :  .  .  . 

212.  When  the  progression  is  decreasing,  and  the  number  of  termg  In- 
finite, what  is  the  expression  for  the  value  of  the  sum  of  the  series  ? 
18 


290  ELEMENTARY       A  L  O  E  B  K  A  . 

containing  an  indefinite  number  of  terms.     In  the  formula, 
„        a  -  Ir 

S  =  T^T' 

substitute  for  I  its  value,  arn~l,  (Art.  208),  and  we  have, 

a  —  arn 


S  = 


1  —  r 


which  expresses  the  sum  of  n  terms  of  the  progression. 
This  may  be  put  under  the  form, 

„  a  ar* 

&  ==  i~ 


1  —  r       I  —  r 
Now,  since  the  progression  is  decreasing,  r  is  a  proper 
fraction  ;  and  r*  is  also  a  fraction,  which  diminishes  as  n 
increases.    Therefore,  the  greater  the  number  of  terms  we 
take,  the  more  will  -  -  X  rn  diminish,  and  consequently, 

-L    ^™*   / 

the  more  will  the  entire  sum  of  all  the  terms  approximate 
to  an  equality  with  the  first  part  of  S,  that  is,  to  -  — 
Finally,  when  n  is  taken  greater  than  any  given  number, 
or  n  —  infinity,   then  -  -  x  r*  will  be  less  than  any 

L   —  ~  T 

given  number,  or  will  become  equal  to  0  ;  and  the  expres- 
t  will  then  represent  the  true  value  of  the  sum 


1    ™"~  7* 

of  all  the  terms  of  the  series.  Whence  we  may  conclude, 
that  the  expression  for  the  sum  of  the  terms  of  a  decreasing 
progression,  in  which  the  number  of  terms  is  infinite,  is, 


that  is,  equal  to  the  first  term,  divided  by  1  minus  the  ratio. 


GEOMETRICAL       P  KOU  KE8SI  ON  .  L".»l 

This  is,  properly  speaking,  the  limit  to  which  the  partial 
Bums  approach,  as  we  take  a  greater  number  of  terms  in  the 
progression.  The  difference  between  these  sums  and  -  —  -  , 
may  be  made  as  small  as  we  please,  but  will  only  become 
nothing  when  the  number  of  terms  is  infinite. 

EXAMPLES. 
1.  Find  the  sum  of 


"We  have,  for  the  expression  of  the  sum  of  the  terms, 

*  =  r^  =  —  i=l  =  *  *& 

~3 

The  error  committed  by  taking  this  expression  for  the 
value  of  the  sum  of  the  n  first  terms,  is  expressed  by 


X  r* 


=  -(-V- 

2\3/ 


1  —  r 
First  take  n  =  5 ;  it  becomes, 

3/l\5_        1  JL_ 

2W    :=  2  .  3*  "   162* 

When  n  =  6,  we  find, 

3/l\6_    _1_       1  1 

2\3/    : =   162  X  3   ==  486' 

q 

Hence,  we  see,  that  the  error  committed  by  taking  -  for 

m 

the  sum  of  a  certain  number  of  terms,  is  less  in  proportion 
as  this  number  is  greater. 


29*3  K  L  K  M  K  X  'C  A  K  Y       A  I.  Ci  E  FJ  K  A  . 

2.  Again,  take  the  progression, 

•  -  -  l-  •-•  —  •—'& 
:  2  :  4  :  8  :  16  :  32  : 


We  have,    S  =  --^—  =  -—  :  =  2. 
I  —  r  \ 


Ans. 


_ 

~~  2 


3.  What  is  the  sum  of  the  progression, 

111  1 

1,     —  ,     -  ,     -  ,     -  ,     &c.,  to  infinity. 
10'     100'     1000'     10000' 

K  -        a  l  -   i1       Av< 

8  -  r=~r  -      ~r     V  Ans- 

10 

a  13.  In  the  several  questions  of  geometrical  progres- 
sion, there  are  five  numbers  to  be  considered  : 

1st.  The  first  term,      .     .    a. 

2d.  The  ratio,    .     .     .     .     r. 

3d.  The  number  of  terms,  n. 

4th.  The  last  term,      .     .      /. 

5th.  The  sum  of  the  terms,  S. 

214.  We  shall  terminate  this  subject  by  solving  this 
problem  : 

To  find  a  mean  proportional  between  any  two  numbers, 
as  m  and  n. 

Denote  the  required  mean  by  x.  We  shall  then  have 
(Art.  197), 

a;2  =      m  x  n  ; 


and  hence,  x    =  \/m  x  n. 


213.  How  many  numbers  are  considered  in  a  gcometrlcnJ  progression? 
What  are  they? 

214.  How  do  you  find  a  mean  proportional  between  two  numbers ? 


GEOMETRICAL      PROGRESSION.  293 

That  is  :  Multiply  the  two  numbers  together,  and  extract  the 
square  root  of  the  product. 

1.  What  is  the  geometrical  mean  between  the  numbers 
2  and  8  ? 

Mean  =   18  x  2  =     /lQ   —  4.     Ans. 


2.  What  is  the  mean  between  4  and  16  ?  Ans.     8. 

3.  What  is  the  mean  between  3  and  27  ?  Ans.     9. 

4.  What  is  the  mean  between  2  and  72  ?  -4ws.  12. 

5.  What  is  the  mean  between  4  and  64  ?  ^1«*.  16. 

therefore,  $40  satisfies  the  enunciation. 


294:  ELEMENTARY       ALGEBRA. 


CHAPTER    X. 

OF      LOGARITHMS. 

215.  THE  nature  and  properties  of  the  logarithms  in 
common  use,  will  be  readily  understood  by  considering 
attentively  the  different  powers  of  the  number  10.  They 
are, 

10°  =   1 

101  1=    10 

102  =    100 

103  =    1000 

104  =    10000 
10s   =    10000* 
&c.,  &c. 

It  is  plain  that  the  exponents  0,  1,  2,  3,  4,  5,  <fcc.,  form  an 
arithmetical  series  of  which  the  common  difference  is  1 ;  and 
that  the  numbers  1,  10,  100,  1000,  10000,  100000,  <fcc.,  form 
a  geometrical  progression  of  which  the  common  ratio  is  1 0. 
The  number  10  is  called  the  base  of  the  system  of  logarithms ; 
and  the  exponents  0,  1,  2,  3,  4,  5,  &c.,  are  the  logarithms  of 

215.  What  relation  exists  between  the  exponents  1,  2,  3,  &c.  ?  How 
are  the  corresponding  numbers  10,  100,  1000?  What  is  the  common 
difference  of  the  exponents  ?  What  is  the  common  ratio  of  the  corre- 
sponding numbers  ?  What  is  the  base  of  the  common  system  of  loga- 
rithms ?  What  are  the  exponents  ?  Of  what  number  is  the  exponent  J 
the  logarithm  ?  The  exponent  2  ?  The  exponent  3  ? 


OF      LOGARITHMS. 

the  numbers  which  are  produced  by  raisiug  10  to  the  powers 
denoted  by  those  exponents. 

216.  If  we  denote  the  logarithm  of  any  number  by  ?n, 
then  the  number  itself  will  be  the  mth  power  of  10  ;  that  is, 
if  we  represent  the  corresponding  number  by  M, 

10m  =  M. 

Thus,  if  we  make  m  =  0,  3/will  be  equal  to  1 ;  if  m  =  1, 
M  will  be  equal  to  10,  <fcc.     Hence, 

The  logarithm  of  a  number  is  the  exponent  of  tJie  power 
to  which  it  is  necessary  to  raise  the  base  of  the  system  in 
order  to  produce  the  number. 

217.  If,  as  before,  10  denotes  the  base  of  the  system 
of  logarithms,  m  any  exponent,  and  M  the  corresponding 
number,  we  shall  then  have, 

10"1  =  Jf,  (1.) 

in  which  m  is  the  logarithm  of  M. 

If  we  take  a  second  exponent  n,  and  let  N  denote  the 
corresponding  number,  AVC  shall  have, 

10"  =  Ar,  (2.) 

in  which  n  is  the  logarithm  of  N~. 

If,  now,  we  multiply  the  first  of  these  equations  by  the 
second,  member  by  member,  we  have, 

10"1  x  10"  =   10Bl+"  =  M  X  N; 

but  since  10  is  the  base  of  the  system,  m  +  n  is  the  loga- 
rithm M  x  .2V;  hence, 

216.  If  we  denote  the  base  of  a  system  by  10,  and  the  exponent  by 
m,  what  will  represent  the  corresponding  number?     What  is  the  logarithm 
of  a  number  ? 

217.  To  what  is  the  sum  of  the  logarithms  of  any  two  numbers  equal  ? 
To  what,  then,  will  the  addition  of  logarithms  correspond  ? 


296          ELEMENT  AKT   ALGEBRA. 

The  sum  of  the  logarithms  of  any  tico  numbers  is  equal 
to  the  logarithm  of  their  product. 

Therefore,  the  addition  of  logarithms  correspond*  to  the 
multiplication  of  their  numbers. 

218.  If  we  divide  Equation  ( 1 )  by  Equation  ( 2 ),  mem- 
ber by  member,  we  have, 

JO1 
10 

but  since  10  is  the  base  of  the  system,  m  —  n  is  the  loga- 
rithm of  -=^.;  hence, 

If  one  number  be  divided  by  another,  the  logarithm  of 
the  quotient  will  be  equal  to  the  logarithm  of  the  dividend, 
diminished  by  that  of  the  divisor. 

Therefore,  the  subtraction  of  logarithms  oorretponA  to 
the  division  of  their  numbers. 

219.  Let  us  examine  further  the  equations, 

10°  =  1 

101  =  10 

102  =  100 
10s  =  1000 
&c.,  &c. 

It  is  plain  that  the  logarithm  of  1  is  0,  and  that  the  loga- 
rithm of  any  number  between  1  and  10,  is  greater  than 


218.  If  one  number  be  divided  by  another,  what  will  the  logarithm 
of  the  quotient  be  equal  to  ?  To  what,  then,  will  the  subtraction  of  loga- 
rithms correspond? 

•J19.  What  is  the  logarithm  of  1  ?  Between  what  limits  are  the  loga- 
rithms of  all  numbers  between  1  and  10?  Row  are  they  generally  ex- 
pressed ? 


OF      LOGARITHMS.  297 

0  and  less  than  1.    The  logarithm  is  generally  expressed  by 
decimal  fractions ;  thus, 

log  2  =  0.301030. 

The  logarithm  of  any  number  greater  than  10  and  less 
than  100,  is  greater  than  1  and  less  than  2,  and  is  expressed 
by  1  and  a  decimal  fraction ;  thus, 

log  50  =   1.698970. 

The  part  of  the  logarithm  which  stands  at  the  left  of  the 
decimal  point,  is  called  the  characteristic  of  the  logarithm. 
The  characteristic  is  always  one  less  than  the  number  of 
places  of  figures  in  the  number  whose  logarithm  is  taken. 

Thus,  in  the  first  case,  for  numbers  between  1  and  10, 
there  is  but  one  place  of  figures,  and  the  characteristic  is  0. 
For  numbers  between  10  and  100,  there  are  two  places  of 
figures,  and  the  characteristic  is  1  ;  and  similarly  for  other 
numbers. 


TABLE     OF    LOGARITHMS. 

22O.  A  table  of  logarithms  is  a  table  in  which  are  writ- 
ten the  logarithms  of  all  numbers  between  1  and  some  other 
given  number.  A  table  showing  the  logarithms  of  the 
numbers  between  1  and  100  is  annexed.  The  numbers  are 
written  in  the  column  designated  by  the  letter  N,  and  the 
logarithms  in  the  column  designated  by  Log. 


How  is  it  with  the  logarithms  of  numbers  between  10  and  100?  What 
is  that  part  of  the  logarithm  called  which  stands  at  the  left  of  the  char- 
acteristic? What  is  the  value  of  the  characteristic? 

22<>.  What  is  a  table  of  logarithms?     Explain  the  manner  of  finding 
the  logarithms  of  numbers  between  1  and  100? 
13* 


298 


ELEMENTARY      ALGEBRA. 


TABLE. 


N. 

Log. 

N. 

Log. 

N. 

Log. 

N. 

Log. 

,1 

2 

3 
4 
5 

0.000000 
0.301030 
0.477121 
0.602060 
0.698970 

26 
27 
28 
29 
30 

1.414973 
1.431364 
1.447158 
1.462398 
1.477121 

51 
52 
53 
54 
55 

1.707570 
1.716003 
1.724276 
1.732394 
1.740363 

7(5 

77 
78 
79 
80 

1.880814 
1.886491 
1.892095 
1.897627 
1.903090 

6 
7 
8 
1 

10 

0.778151 
0.845098 
0.903090 
0.954243 
1.000000 

31 
32 
33 
34 

35 

1.491362 
1.505150 
1.518514 
1.531479 

1.544068 

56 
57 

58 
59 
60 

1.748188 
1.755875 
1.763428 
1.770852 
1.778151 

81 

82 
83 
84 
85 

1.908485 
1.913814 
1.919078 
1.924279 
1.929419 

11 
12 
13 

14 
15 

1.041393 
1.079181 
1.113943 
1.146128 
1.176091 

36 
37 
38 
39 
40 

1.556303 
1.568202 
1.579784 
1.591065 
1.602060 

61 
62 
63 
64 

65 

1.785330 
1.792392 
1.799341 
1.806180 
1.812913 

86 
87 
88 
89 
90 

1.934498 
1.939519 
1.944483 
1.949390 
1  .  954243 

16 
17 
18 
19 
20 

1.204120 
1.230449 
1.255273 
1.278754 
1.301030 

41 
42 
43 
44 
45 

1.621784 
1.623249 
1.633468 
1.643453 
1.653213 

66 
67 
68 
C9 
70 

1.819544 
1.826075 
1  .  832509 
1.838849 
1  .  845098 

91 
92 
93 
94 
95 

1.959041 
1.963788 
1.968483 
1.973128 
1.977724 

21 
22 
23 
24 

25 

1.322219 
1  .  342423 
1.361728 
1.380211 
1.3979-10 

46 
47 

48 
49 
'  50 

1.662758 
1.672098 
1.681241 
1.690196 
1.698970 

71 
72 
73 
74 
75 

1.851258 
1  .  857333 
1.863323 
1.869232 
1.875061 

96 
97 
98 
99 

100 

1.982271 
1.986772 
1.991226 
1.995635 
2.000000 

EXAMPLES. 

1 .  Let  it  be  required  to  multiply  8  by  9,  by  means  of 
logarithms.  We  have  seen,  Art.  216,  that  the  sum  of  the 
logarithms  is  equal  to  the  logarithm  of  the  product.  There- 
fore, find  the  logarithm  of  8  from  the  table,  which  is  0.903090, 
ami  then  the  logarithm  of  9,  which  is  0.954243 ;  and  their 
sum,  which  is  1.857333,  will  be  the  logarithm  of  the  product. 
In  searching  along  in  the  table,  we  find  that  72  stands  oppo- 
site this  logarithm ;  hence,  72  is  the  product  of  8  by  9. 


OF      LOGABITI1MS. 


299 


2.  What  is  the  product  of  V  by  12? 

Logarithm  of    7  is, 

Logarithm  of  12  is,       .        .        . 

Logarithm  of  their  product, 
and  the  corresponding  number  is  84. 

3.  What  is  the  product  of  9  by  11  ? 

Logarithm  of    9  is, 
Logarithm  of  11  is, 

Logarithm  of  their  product, 
and  the  corresponding  number  is  99. 


0.845098 
1.079181 


1.924279 


0.954243 
1.041393 


1.995636 


4.  Let  it  be  required  to  divide  84  by  3.    We  have  seen 
in  Art.  218,  that  the  subtraction  of  Logarithms  corresponds 
to  the  division  of  their  numbers.     Hence,  if  we  find  the 
logarithm  of  84,  and  then  subtract  from  it  the  logarithm  of 
3,  the  remainder  will  be  the  logarithm  of  the  quotient. 

The  logarithm  of  84  is, 
The  logarithm  of    3  is, 

Their  difference  is,         ... 
and  the  corresponding  number  is  28. 

5.  What  is  the  product  of  6  by  7? 

Logarithm  of  6  is, 
Logarithm  of  7  is, 

Their  sum  is, 1.623249 

and  the  corresponding  number  of  the  table,  42. 


1.924279 
0.477121 

1.447158 


0.778151 
0.845098 


RECOMMENDATIONS  OF  DAVIES'  MATHEMATICS. 


DAVIKS'  COUBKE  OF  MATHEMATICS  are  the  prominent  Texl-Jiooks  in  ntott 
of  the  Golltgts  of  the  United  Sltite.i,  and  also^in  the  various  Scho<  is  in.  I 
Acitdttinics  throughout  the  Union. 

YORK,  PA.,  Auy.'2\  !«•>. 

Davits'  Seriet  of  Mathematict  1  deem  the  very  best  I  ever  saw.  From  a  nunihei 
i»f  authors  I  selected  it,  after  a  careful  perusal,  as  a  course  of  stndy  to  be  pursued  by 
Hie  Teachers  attending  the  sessions  of  the  Turk  Co.  Normal  School— believing  It  also 
n  b.-  well  adapted  to  the  wants  of  the  schools  throughout  our  country.  Already  r.vo 
»  indred  schools  are  supplied  with  DAVI&'  valuable  Sit-ins  of  Arithmetic* ;  and  I 
Mlv  believe  that  in  »  very  short  time  the  Teachers  of  our  country  en  ma»*e  will  i* 
i  .rs.'ed  in  imparting  instruction  throngh  the  medium  of  this  new  and  easy  mett-v- 
1  .-itijtlysls  of  numbers.  A.  E.  BLAIR, 

Principal  of  York  Co.  Normal  Scb»m 

JACKSON  UNION  SCHOOL.  MICIIIGAX.  Xei>t.  2.\  K,-. 

Mrc*i:8  A.  S.  BARSRS  &  Co. :— I  take  pleasure  in  adding  my  testimony  in  favor  ol 
I'.irifx'  y/n-ifg  of  Mathfmativ*,  ax  published  by  yon.  We"  have  used  these  work>  in 
this  school  for  more  than  four  years  ;  a»d  «i  «v)l  satisfied  are  we  of  their  nperioHtl 
over  any  other  Series,  that  we 'neither  contemplate  making,  iior  de-ire  to  make,  HHJ 
change 'in  that  direction.  Yours  truly,  E.  L.  IJIl'LKY. 

NKW  UKITAIK.  June  I'M.  1V.S. 

MI-.SHKS.  A.  S.  P,AI:XI:S  &  Co . :—  I  h:ive  eNamincd  I><irl-  x'  Sf,  -<V«  <if  .t>-if!i.- ,  '" 
with  some  cure.  They  appear  well  adapted  for  the  different  <rra;lcs  of  ><h<>o]>  t'ur 
which  they  aro  designed.  The  language  is  clear  and  precise :  e.cch  principle  i* 
thoroughly  analyzed,  and  the  whole  so  arranged  as  to  facilitate  the  work  of  instruc- 
tion. Having  observed  the  satisfaction  and  success  with  which  the  different  ho,.k> 
have  been  used  by  eminent  tenchers,  it  srivc,*  me  pleasure  to  commend  them  to  others. 
DAVID  N.  C.\Ml\f'riii(.-ij><il<>fConn.  State  Normal  Scho»'. 

I  have  long  regarded  Davits'  Series  of  Mathematical  Tert-Books  as  far  superioi 
to  any  now  before  the  public.  We  find  them  in  every  way  adapted  to  the  wants  n'. 
the  Normal  School,  and  we  use  no  other.  A  unity  of  sys'tem  and  method  runs  through- 
out the  series,  and  constitutes  one  of  its  preat  excellences.  Especially  in  the  Arith- 
metics the  author  has  earnestly  endeavored  to  supply  the  wants  of  our  Common  and 
Union  Schools:  and  his  success  is  complete,  and  undeniable.  1  know  of  no  Arith- 
metics which  exhibit  so  clearly  the  philosophy  of  numbers,  and  at  the  same  time  lead 
the  pupil  surely  on  to  readiness  and  practice.  A.  8.  WELCH. 

From  PBOF.  G.  W.  PLVMPTON,  late  of  the  Sbite  Normal  School,  X.  Y. 
Out  of  a  great  number  of  Arithmetics  that  I  have  examined  during  the  past  year,  1 
find  none  that  will  compare  with  Datie/f  Intellectual  and  Dar.iej  Analytical  and 
Practical  Arithmetics,  in  clearness  of  demonstration  or  philosophical  arrangement. 
I  shall  with  pleasure  recommend  the  use  of  these  two  excellent  works  to  those  who 
go  from  our  institution  to  teach. 

From  C.  MAY,  JK.,  School  Commissioner,  Keene,  2V.  H. 

I  bare  carefully  examined  Da-tie*'  Seriei  of  Arithmetics,  and  Higher  Mat'if- 
tnaticit,  and  am  prepared  to  say  that  I  consider  them  far  superior  to  any  with  which 
I  am  acquainted. 

ffiim  Joim  L  CAMPBBLL,  Professor  uf  Jfitt?ifmatic*,  Natural  PhUovophy,  an/I 
Astronomy,  in  Wabash  College,  Iwlidna. 

W  ABASH  COLI.KOK.  Jntif  22.  !-.> 

M«SRS.  A.  S.  BAKHES  &  Co. : — GF.NTLKMKN  :  Every  text-book  on  Science  pvo|"  r'y 
•<  IINJSIS  of  two  jiarts — the  philosophical  and  the  illuxtratitt.  A  proper  cointiitiiiior. 
.•  -i  pstract  reasoning  and  practical  illustration  is  the  chief  excellence  in  Prof.  l)»vi»- 
Malhcmatical  Works.  I  prefer  his  Arithmetics,  Algebras.  Geometry,  and  Trisronoii: 
•try.  to  all  others  now  in  use.  and  cordially  recommend  them  to  all  who  d<-Miv  i> ». 
vhancemcnt  of  sound  learning,  Yours,  very  truly,  JOHN  L.  CAMPBELL 

PK.IKKSSORS  MAH.SN,  BAHTLETT,  and  CtirROii.  of  the  United  States  Military  Aca.l<-n.» 

West  1'oint.  HfvtDnvfa?  L'niei-ryiti/  Arithmetic: — 

"  In  the  distinctness  wi^h  whieh  the  various  definitions  are  given,  the  clear  «nd 
strictly  mathematical  demonstration  of  the  rules,  the  convenient  form  and  well-chosen 
matter  of  the  tables,  as  well  as  in  the  complete  and  much-desired  application  of  all  to 
the  business  of  the  country,  th*>  Uv>rf»ity  Arithmetic  of  Prof.  Davits  is  st  perior  to 
*!!}•  other  work  of  the  kind  wi;h  •«  h'.-h  '  c  »rp  sc'jnainti-.i  " 


RECOMMENDATIONS 

OF 

PARKER  &  WATSON'S  READERS 


from  PROF.  FREDERICK  S.  JKWKLI.,  of  the  _\><c  York.  State  Jiormal  School 
It  gives  me  pleasure  to  find  in  the  National  Series  at  School  Readers  ample  i.  )i> 
&>r  commendation.     From  n  brief  examination  of  them.  I  urn  led  to  believe  tin.'  <ri 
have  none  equal  to  them.    I  hope  they  will  prove  as  popular  as  they  are  excellent 

f>-nm  IIox.  TiiKonoRK  FRKUXGIIUYSKN,  President  of  Uutgeri?  College,  2f  J. 
.\  cursory  examination  leads  me  to  the  conclusion  that  the  system  contained  ID 
>.li«>.»e  volumes  deserves  the  patronage  of  our  schools,  and  I  have  no  doubt  that  il  w!t 
W-come  extensively  used  in  the  education  of  children  and  youth. 

fnan  N.  A.  HAMILTON,  President  of  Teacher*'  Union,  Whitewater,  n'i*. 

Tin-  National  Readers  and  Speller  I  have  examined,  and  carefully  compared  witb 

others,  and  must  pronounce   them  decidedly  superior,  in   respect  to  literary  merit, 

style.  :in<l  pri.  e.    The  gradation  is  more  complete,  and  the  series  much  more  Owtr»bl« 

for  use  in  our  schools  than  Sanders'  or  McGuffey's. 

f'nnn  PROF.  T.  F.  THICKS-IT  s,  Principal  of  Academy  and  Normal  Scliool, 

Meadrille,  Pa. 

1  am  much  p'eased  with  the  National  Series  of  Readers  after  having  canvassed 
tlieir  merits  pretty  thoroughly.  The  first  of  the  series  especially  pleases  me,  because 
it  affords  the  means  of  teaching  the  ••  word-method"  in  an  appropriate  and  natural 
manner.  They  a'l  arc  progressive,  tbe  rules  of  elocution  are  stated  with  clearness, 
and  the  selection  of  pieces  is  such  as  to  pk-itJc  at  the  s:iine  time  that  they  in*triirt 

Fiom  J.  \V.  SciiF.iiMERiioKN,  A.  15.,  Principal  Coll.  Institute,  Stiddletown,  S.  J 
I  consider  tliem  emphatically  the  Readers  of  the  present  day,  and  I  believe  thu 
,heir  ii.rrinsic  merits  will  insure  for  them  a  full  measure  of  popularity. 

From  PETER  Rouoicr,  Principal  Public  School  Xo.  10,  Brooklyn. 
It  gives  me  great  pleasure  to  be  able  to  bear  my  unqualified  testimony  to  the  >-xcel 
lence  of  the  National  Series  of  Readers,  by  PAKKF.K  and  WATSON.  The  gradation  of 
the  books  of  the  series  is  very  fine  :  we  have  reading  in  its  elements  and  in  its  highest 
style.  The  fine  taste  displayed  in  the  selections  ami  in  the  collocation  of  the  piece? 
it'serves  much  praise.  A  distinguishing  feature  of  the  series  is  the  variety  of  thf 
subject-matter  and  of  the  style.  The  practical  teacher  knows  the  value  of  this  charm: 
terisik  for  the  development  of  the  voice.  The  authors  seein  to  have  kept  constantly 
in  view  tlie  fact  that  a  reading-book  is  designed  for  children,  and  therefore  they  have 
surceeded  in  forming  a  very  interesting  and  improving  collection  of  reading-matter, 
highly  adapted  to  the  wants  and  purposes  of  the  school-room.  In  short,  I  look  upon 
the  National  Scries  of  ReaUers  as  a  great  success. 

from  \.  P.  HARRINGTON,  Principal  of  Union  School,  MaraUion,  N.  }'. 
These  Renders,  in  my  opinion,  aie  the  best  I  have  ever  examined.  The  rhetorical 
exercises,  in  particular,  are  superior  Ui  any  thing  of  the  kind  I  have  ever  seen.  I  have 
iiad  tn-tter  success  with  my  reading  classes  since  I  commenced  training  them  i.n  those 
ilnii  1  ever  met  with  before.  The  marked  vowels  in  the  reading  exori-ises  coiury  10 
'.lie  reaiicr's  iniiid  ut  once  the  astonishing  f«ct  that  he  has  been  accustomed  to  im-pro- 
:.»nnve  more  than  one-third  of  the  words  of  the  English  language. 

f"rrm  OriAr.i.KS  S.  HALSF.V.  Prinrijml  C<>Ufyi<itt  Inntitute,  Nnrtun.  A".  J. 

In  ih<  simplicity  and  clearness  with  which  the  principles  are  stated,  in  the  appro 
•r;ati-n<*s  of  the  selections  for  reading,  and  i:i  the  happy  adaptation  of  the  dilf-ient 
-.•rt.«  V,  ihe  series  to  oach  .-ther.  these  works  are  s-nperior  to  any  other  text  books  on 

i ,  si  tiji-ct  which  I  have  examined. 

t'ntin  WILLIAM  TRAVIS,  Principal  of  Union  School,  flint,  J/icA. 
i  b    .v  exan'ined  the  Nc.tiona!  Series  of  Readers,  and  am  delighted  to  find  'l  so  fat 
.     •,   ani'H  of  most  other  serii-s  now  in  use,  and  so  wtll  adapted  to  the  wains  ..f  ik» 
i     )  .:    Srhoo;».     It   is  in, equaled   in   the  skillful  arrangement  of  the  m.-iti-rihl  r.><  it 
V-ij  ii'u!  typiwriipby.  and  the  general  neat  and  inviting  appearance  of  it*  several 
!H>.  I.*     I  predict  tor  it  a  cordial  Ve'.coine  and  a  general  introduction  by  many  of  0111 
nv>st  *t,tei(.]  isinj  ton  hers. 


RECOMMENDATIONS 


CLARK'S  ENGLISH  GRAMMAR. 

We  cannot  better  set  forth  the  merits  of  this  work  than  by  quoting  a  part  of  a  com- 
munication from  Prot  F.  8.  JEWKLL,  of  tlie  New  York  State  Normal  School,  in  wblct 
school  this  Grammar  is  now  used  as  the  text  book  on  this  subject : — 

•;  CT.ARK'S  SYSTKM  or  GRAMMAR  is  worthy  of  the  marked  attention  of  the  friends  01 
jilucntion.  Its  points  of  excellence  are  of  the  most  decided  character,  and  will  nc*. 
*>'>n  be  surpassed.  Among  them  are — 

1st.  '-The  justness  of  its  around  principle  of  classification.    There  is  no  simple,  phi!- 

isophieal.  and  practical  classification  of  the  elements  of  language,  other  than  thnt  bui't 

;,  iheir  use  or  i;fltce.     Our  tendencias  hitherto  to  follow  tho  analogies  of  the  classical 

.«!!gu.-is:«-s.  and  classify  extensively  according  to  forms,  have  been  mischievous  and  ab- 

minl.     It  is  time  we  corrected  them. 

2d.  '•  Its  thorough  and  yt  simple  and  transparent  analysis  of  the  elements  of  the 
language  according  to  its  ground  principle.  Without  such  an  analysis,  no  broad  and 
comprehensive  view  of  the  structure  and  power  of  the  language  can  be  attained.  The 
al.setx-e  of  this  analysis  has  hitherto  precipitated  the  study  of  Grammar  upon  a  surface 
of  dry  details  and  bare  authorities,  and  useless  technicalities. 

3d." "  Its  happy  method  of  illustrating  the  relations  of  elements  by  diagrams.  These, 
however  uncouth  they  may  appear  to  the  novice,  sre  really  simple  and  philosophical. 
Of  their  utility  there  can  be  no  question.  It  is  supported  by  the  usage  of  other  sci- 
ences, and  has  bc«-n  demonstrated  by  experience  in  this. 

4th  "The  tenden'-y  of  the  system,  when  rightly  taught  and  faithfully  carried  out, 
to  cultivate  habits  of  nice  discrimination  and  close  reasoning,  together  with  skill  in 
illustrating  truth.  In  this  it  is  not  excelled  by  any,  unless  it  be  the  mathematical  sci- 
ences, and  even  there  it  hn>  this  advantage.  th:it  it  deals  with  elements  more  within 
tire  present  gra>p  of  the  intellect  On  this  point  I  speak  advisedly. 

5th.  -'The  system  is  thoroughly  progressive  and  practical,  and  as  such.  American  in 
[•f  character.  "It  does  not  adhere  to  old  usages,  merely  because  thev  are  veneraliy 
musty;  anil  yet  it  does  ma  discard  things  merely  because  they  are  old.  or  are  in  un- 
Importxni  minutie  not  prudishly  perfect.  It  does  nut  overlook  details  and  technicali- 
ties, nor  does  it  allow  them  to  interfere  with  plain  philosophy  or  practical  utility. 

'•  Let  any  clear-headed,  independent-minded  teacher  master  the  system,  and  then 
give  it  a  fa'ir  trial,  and  there.wiil  be  no  doubt  as  to  his  testimony." 

A   Testimonial  from  tlie  Pr  in  ci juris  of  the  Public  School*  of  Rochester,  N.  Y. 
We  reg.-ird  CLARK'S  GRAMMAR  as  the  clearest  in  its  analysis,  the  most  natural  and 

logical  itt  its  arrangement,  the  most  concise  and  accurate  in  its  definitions,  the  mos-. 

M  .-lemittic  in  design,  and  the  best  adapted  to  the  use  of  schools  of  any  Grammar  »tth 

which  we  are  acquainted. 

C  C.  MKSKRVE,  WM.  C.  FEGLES. 

M  D.  ROWLEY,  OHN  ATWATKK, 

C.  R.  BUBRICK,  EDWARD  WEBSTER, 

J.  R.  VOSBURG,  8.  W.  STARKWEATHER, 

E.  K.  ARMSTRONG  PHILIP  CURTISS. 

LAWEF.XCR  INSTITUTE,  Brooklyn,  Jan  15, 1859. 

MESSRS.  A.  S.  BARXFS  &  Co: — Having  used  Clark's  New  Grammar  since  Its  publica- 
tion, i  do  most  unhesitatingly  recommend  it  as  a  work  of  cupvrior  merit  By  the  ut* 
of  ii-i  other  work,  ami  I  have  used  several,  have  I  been  enabled  to  advance  uiy  pupil* 
v.  njiidly  and  thoroughly. 

The  author  has,  by  an  "Etymological  Chart  and  a  system  of  Diagrams,  made  Gram 
:iar  the  study  that  it  ought  to  be,  interesting  as  well  as  nseful. 

MARGARET  S.  LAWRENCE,  Principal. 


WELCH'S  ENGLISH  SENTENCE, 

frtttK  P*or.  J.  R  BOISE,  A.  M.,  Profetxor  of  the  Latin  and  Greek  Langvaget  etna 

Literature  in  the  University  of  Michigan. 

Tliis  work  belongs  to  a  new  era  in  the  grammatical  study  of  our  own  language.  \V  r 
n»/.Mr<l  nothing.  In  expressing  the  opinion,  that  for  severe,  searching,  and  ~ezlianstiv« 
m»!ysis.  the  work  of  1'rofessor  Welch  is  second  to  none.  His  book  Is  not  Intended  fo« 
beginners,  but  only  f--r  advanced  students,  snd  by  inch  only  It  will  be  understood  tnl 
appreciated. 


MONTEITH  AND  PIcNALLY'S  GEOGRAPHIES 

THE  MOST  SUCCESSFUL  SERIES  EVER  ISSUED. 


RECOMMENDATIONS. 

A.  B.  CLARK,  Principal  of  one  of  the  largest  Public  Schools  In  Brooklyn  (*yt:— 
M  have  used  over  a  thousand  copies  of  Monteith's  Manual  of  Geography  sine*  Iti 
adoption  by  :he  Board  of  Education,  and  am  prepared  to  say  it  is  the  beat  wuik  fo» 
Junior  and  intermediate  classes  in  our  schools  1  have  ever  seen." 


The  Series,  in  whole  or  in  part.  hrt«  I/fen  adopted  in  th« 


New  York  State  Normal  School. 
New  York  City  Normal  School. 
New  .Jersey  Stale  Normal  Si-hool. 
Kentucky  State  Normal  School. 
Imlimia  State  Normal  School. 
Ohio  ^rate  Normal  School. 
Michigan  Stale  Normal  School. 
York  County  (Pa.)  Normal  School. 
Brooklyn  1'olytechnio,  Institute. 
Cleveland  Female  Seminary. 
Public  Schools  of  MilwMiikie. 
Public  Schools  of  Pittsburgh 
Public  Schools  of  Lancaster,  Pa. 
Public  Schools  of  New  Orleans. 


Public  Schools  of  New  York. 
Public  Schools  of  Brooklyn,  L.  I. 
Public  Schools  of  New  liaven. 
Public  Schools*  of  Toledo,  Ohio. 
Public  Schools  of  Norwalk,  Conn. 
Public  Schools  of  Richmond.  Va. 
Public  Schools  of  Madison,  Wig. 
Public  Schools  of  Indiananc'is. 
Public  Schools  of  Springfield,  Maw. 
Public  Schools  of  Columbus.  Ohio. 
Public  Schools  of  Hartford.  Conn. 
Public  Schools  of  Cleveland,  Ohio. 

And  other  places  too  numerous  to 
mention. 


They  have  also  been  recommended  by  the  State  Superintendents  of  ILLINOIS, 
INDIANA,  WISCONSIN.  Missouiti,  Nor.ru  CAROLINA,  ALABAMA,  and  by  numerous 
Teachers'  Associations  and  Institutes  throughout  the  country,  ami  are  in  successful 
use  in  a  multitude  of  Public  and  Private  Schools  throughout  the  United  States. 


From  PROF.  WM.  F.  PIIELPS,  A.  M.,  Principal  ofth«  New  Jersey  Staff 
jformal  School. 

TRKNTOK.  Junf  17.  iSfH. 

MF.SRXS.  A.  S.  BARNKS  «fc  Co.:  —  GBNTLF.MKN:  It  gives  me  much  pleasure  to  state 
th.tt  MrNally's  Geography  has  been  used  in  this  Institution  from  its  organization  in 
1SS5.  with  great  acceptance.  The  author  of  this  work  has  avoided  on  one  hand  the 
extreme  of  being  too  meager,  and  on  the  other  of  going  too  much  into  detail,  while 
he  has  presented,  in  a  clear  and  concise  manner,  all  those  leading  facts  of  Descriptive 
Geography  which  it  is  important  for  the  young  to  know.  The  mtps  are  accurate  and 
well  executed,  the  type  clear,  and  indeed  the  entire  work  Is  a  decided  success.  I  most 
cheerfully  commend  it  to  the  profession  throughout  the  country. 

Very  *ruly  yours,  WM.  F.  PHELP8. 

From  W.  V.  DAVIS,  Piindpnl  of  High  School,  Lancaster,  Pa. 

LANCASTER,  PA.,  Junf  26.  1S58. 

DKAR  SIRS:—  I  have  examined  yonr  National  Geographical  S<*rien  with  much 
care,  and  find  them  most  excellent  works  of  their  kind.  They  have  been  used  in  the 
various  Public  Schools  of  this  city,  ever  since  their  publication,  with  treat  succee*  and 
satisfaction  to  both  pupil  and  teacher.  All  the  Geographies  embraced  in  your  series 
are  weil  adapted  to  school  purposes,  and  admirably  calculated  to  impart  to  the  pupil, 
in  a  very  attractive  manner,  a  complete  knowledge  of  a  science,  annually  becoming 
more  useful  and  important  Their  maps,  illustrations,  mid  typography,  are  unsur- 
passed. One  peculiar  feature  of  McNally's  Geography  —  and  which  will  recommeud 
it.  at  once  to  every  practical  teacher  —  is  the  arrangement  of  i»s  maps  and  lessons; 
each  map  fronts  the  particular  lesxm  which  it  is  designed  to  illustrate  —  thus  enabling 
the  scholar  to  prepare  his  ta>k  without  that  constant  turning  over  of  leave*,  or  refrr- 
•nce  to  a  separate  book,  as  Is  necessary  with  most  othor  Geographies.  Yours.  &c. 

Messrs.  A.  S.  BAKNKS  A  Co.,  New  York.  V.  W.  DAVIS. 

From  CHARLES  BARNRS,  late  Preaiilfnt  State  Ttavktr*'  Amouiutiim,  and  Stijifhi- 
tendent  qfth,«  Public  Softools  at  Neic  Albany,  Indiana. 

MKSSHS.  A.  S.  BAKNES  <fe  Co.:  —  DK.AK  SIRS:  I  have  examined  with  eonrider:ib'« 
care  the  Series  of  Geographies  published  by  you,  and  have  no  hesitktion  in  saying 
that  it  Is  altogether  the  best  with  wllch  I  am  acquainted.  A  trial  of  more  than  • 


in  ih«  Public  School 


ools  .if  this  city  has  demonstrated  that  CornfU  is  utterlv  unflt 
Tours'.  Ac.  C.  BARNES. 


RECOMMENDATIONS 

or 
MONTEITH'S  HISTORY  OF  THE  UNITED  STATES. 


Fhis  volume  is  designed  for  youth,  and  we  think  the  author  has  been  unusual  / 
»'iccessfiil  in  its  arrangement  and  entire  preparation.  Books  of  the  same  design  a  i 
too  often  beyond  the  full  understanding  of  the  scholar.  As  history  is  so  much  no-,,- 
lected  in  all  our  schools,  the  publication  of  such  a  work  as  this  should  be  hailed  wiih 
pleasure;  for  if  scholars  find  their  first  studies  of  history  pleasant,  it  will  become  • 
pleasure  rather  than  a  task.  This  is  a  book  of  SS  pages,  and  finely  illustrated.  It  i>  in 
every  way  worthy  of  a  place  in  every  Public  School  in  the  State. — Maine  Tettcker. 

This  is  a  most  capital  work  :  just  the  thing  for  children.  Our  boy  commenced  ,he 
"tudy  of  it  the  day  it  came  to  hand.  It  is  arranged  in  the  catechetical  form,  and  is 
fiiK-ly  illustrated  with  maps,  with  special  reference  to  the  matter  discussed  in  the  text, 
It  begins  with  the  first  discoveries  of  America.  »nd  comes  down  to  the  laying  of  the 
Atlantic  Telegrnph  Cable.  Many  spirited  engravings  are  given  to  illustrate  the  work. 
It  also  contains  brief  Biographies  of  all  prominent  men  who  have  identified  them- 
selves  with  the  history  of  this  country.  It  is  the  best  work  of  the  kind  we  have 
Been.— Cluster  County  Times. 


WILLARD'S   HISTORIES. 

from  RET.  HOWARD  MALCOLM,  D.  D.,  President  <tf  th«  University  of  Leirisburg. 

I  have  examined,  during  the  thirteen  years  that  I  have  had  charge  of  a  College, 
many  School  Histories  of  the  United  States,  and  have  found  none,  on  the  whole,  so 
proper  for  n  text-book  as  that  of  Mrs.  Willard.  It  is  neither  too  short  nor  too  long, 
«11  the  space  given  to  periods,  events,  and  persons,  is  happily  proportioned  to  their 
importance.  The  style  is  attractive  and  lucid,  and  the  narrative  so  woven,  as  both 
to  sustain  the  interest  and  aid  the  memory  of  the  student.  Candor,  impartiality,  and 
accuracy,  are  conspicuous  throughout.  I  think  no  teacher  intending  to  commence  a 
history  class  will  be  disappointed  in  adopting  this  book. 

MBS.  L.  H.  SIGOUBNKT,  Vt4  diilinguithtd  Authoress,  writes: 

Mrs.  Willard  should  be  considered  as  a  benefactress  not  only  by  her  own  sex,  of 
whom  she  became  in  early  years  a  prominent  and  permanent  educator,  but  by  the 
country  at  large,  to  whose  good  she  has  dedicated  the  gathered  learning  and  faithful 
labor  of  life's  later  periods.  The  truths  that  she  has  recorded,  and  the  principles  that 
she  has  impressed,  will  win,  from  a  future  race,  gratitude  that  cannot  grow  old,  aud  a 
garland  that  will  never  fade. 

DANIEL  WEBSTBR  wrote,  in  a  letter  to  the  Author: 

I  cannot  better  express  my  sense  of  the  value  of  your  History  of  the  United  Statf-t, 
than  by  saying  I  ieep  it  near  me  as  a  book  of  reference,  accurate  in  fact*  and  dates. 


DWIGHT'S   MYTHOLOGY. 

The  mythology  of  the  Grecians  and  Romans  Is  so  closely  interlinked  with  the  his- 
•  TV  »nd  literature  of  the  world,  thtt  some  knowledge  ot  It  is  indispensable  to  any 
,-nn!arly  familiarity  with  either  that  history  or  literature.  We  have  seon  no  book  so 
!••>..  vrnu-nt  in  (size  that  contains  so  full  and  elegant  an  exposition  of  mythology  ax  thi 
one  before  us.  It  will  be  found  at  once  a  most  interesting  and  a  most  useful  book  to 
any  one  who  wiskos  an  acquaintance  with  the  splendid  myths  and  fables  with  whiob 
the  great  masters  of  ancient  learning  amused  their  leisure  and  cheated  tbtir  faith  - 
tficMgan  Journal  offduca'ien.  , 


UCSB  LIBRARY 


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